A number of K balls are dropped one by one from the root of a fully binary tree structure FBT. Each time the ball being dropped first visits a non-terminal node. It then keeps moving down, either follows the path of the left subtree, or follows the path of the right subtree, until it stops at one of the leaf nodes of FBT. To determine a ball’s moving direction a flag is set up in every non-terminal node with two values, either false or true. Initially, all of the flags are false. When visiting a non-terminal node if the flag’s current value at this node is false, then the ball will first switch this flag’s value, i.e., from the false to the true, and then follow the left subtree of this node to keep moving down. Otherwise, it will also switch this flag’s value, i.e., from the true to the false, but will follow the right subtree of this node to keep moving down. Furthermore, all nodes of FBT are sequentially numbered, starting at 1 with nodes on depth 1, and then those on depth 2, and so on. Nodes on any depth are numbered from left to right.

For example, Fig. 1 represents a fully binary tree of maximum depth 4 with the node numbers 1, 2, 3, ..., 15. Since all of the flags are initially set to be false, the first ball being dropped will switch flag’s values at node 1, node 2, and node 4 before it finally stops at position 8. The second ball being dropped will switch flag’s values at node 1, node 3, and node 6, and stop at position 12. Obviously, the third ball being dropped will switch flag’s values at node 1, node 2, and node 5 before it stops at position 10.

Now consider a number of test cases where two values will be given for each test. The first value is D, the maximum depth of FBT, and the second one is I, the I-th ball being dropped. You may assume the value of I will not exceed the total number of leaf nodes for the given FBT.

Please write a program to determine the stop position P for each test case.

For each test cases the range of two parameters D and I is as below:

2 ≤ D ≤ 20, and 1 ≤ I ≤ 524288.

Input

Contains l + 2 lines.

Line 1 l the number of test cases

Line 2 D1 I1 test case #1, two decimal numbers that are separated by one blank

...

Line k + 1 Dk Ik test case #k

Line l + 1 Dl Il test case #l

Line l + 2 -1 a constant ‘-1’ representing the end of the input file

Output

Contains l lines.

Line 1 the stop position P for the test case #1

...

Line k the stop position P for the test case #k

...

Line l the stop position P for the test case #l

Sample Input

5
4 2
3 4
10 1
2 2
8 128
-1
Sample Output

12

7

512

3

255

题目类型：二叉树查找

算法分析：由于每个节点具有开关属性，则当标号k为奇数时其为走左子树的第(k+1)/2个节点(也即其在左子树中的顺序标号)，否则为走右子树的第k/2个节点，一共走D-1次

You have n boxes in a line on the table numbered 1 . . . n from left to right. Your task is to simulate 4 kinds of commands:

• 1 X Y : move box X to the left to Y (ignore this if X is already the left of Y )

• 2 X Y : move box X to the right to Y (ignore this if X is already the right of Y )

• 3 X Y : swap box X and Y

• 4: reverse the whole line.

Commands are guaranteed to be valid, i.e. X will be not equal to Y .

For example, if n = 6, after executing 1 1 4, the line becomes 2 3 1 4 5 6. Then after executing 2 3 5, the line becomes 2 1 4 5 3 6. Then after executing 3 1 6, the line becomes 2 6 4 5 3 1. Then after executing 4, then line becomes 1 3 5 4 6 2

Input

There will be at most 10 test cases. Each test case begins with a line containing 2 integers n, m (1 ≤ n, m ≤ 100, 000). Each of the following m lines contain a command.

Output

For each test case, print the sum of numbers at odd-indexed positions. Positions are numbered 1 to n from left to right.

Sample Input

6 4
1 1 4
2 3 5
3 1 6
4
6 3
1 1 4
2 3 5
3 1 6
100000 1
4

Sample Output

Case 1: 12

Case 2: 9

Case 3: 2500050000

题目类型：双重链表

算法分析：使用left[x]表示x的左边节点标号，right[x]表示x的右边节点标号，之后按照题目模拟即可。注意答案可能超出int的范围

Suppose you have to evaluate an expression like A*B*C*D*E where A,B,C,D and E are matrices. Since matrix multiplication is associative, the order in which multiplications are performed is arbitrary. However, the number of elementary multiplications needed strongly depends on the evaluation order you choose.

For example, let A be a 50*10 matrix, B a 10*20 matrix and C a 20*5 matrix. There are two different strategies to compute A*B*C, namely (A*B)*C and A*(B*C).

The first one takes 15000 elementary multiplications, but the second one only 3500.

Your job is to write a program that determines the number of elementary multiplications needed for a given evaluation strategy.

Input

Input consists of two parts: a list of matrices and a list of expressions.

The first line of the input file contains one integer n (1 ≤ n ≤ 26), representing the number of matrices in the first part. The next n lines each contain one capital letter, specifying the name of the matrix, and two integers, specifying the number of rows and columns of the matrix.

The second part of the input file strictly adheres to the following syntax (given in EBNF):

SecondPart = Line { Line }

Line = Expression

Expression = Matrix | "(" Expression Expression ")"

Matrix = "A" | "B" | "C" | ... | "X" | "Y" | "Z"

Output

For each expression found in the second part of the input file, print one line containing the word ‘error’ if evaluation of the expression leads to an error due to non-matching matrices. Otherwise print one line containing the number of elementary multiplications needed to evaluate the expression in the way specified by the parentheses.

Sample Input

9
A 50 10
B 10 20
C 20 5
D 30 35
E 35 15
F 15 5
G 5 10
H 10 20
I 20 25
A
B
C
(AA)
(AB)
(AC)
(A(BC))
((AB)C)
(((((DE)F)G)H)I)
(D(E(F(G(HI)))))
((D(EF))((GH)I))

Sample Output

0

0

0

error

10000

error

3500

15000

40500

47500

15125

题目类型：表达式计算(栈模拟)

算法分析：使用栈来模拟计算矩阵表达式

There is a famous railway station in PopPush City. Country there is incredibly hilly. The station was built in last century. Unfortunately, funds were extremely limited that time. It was possible to establish only a surface track. Moreover, it turned out that the station could be only a dead-end one (see picture) and due to lack of available space it could have only one track.

The local tradition is that every train arriving from the direction A continues in the direction B with coaches reorganized in some way. Assume that the train arriving from the direction A has N ≤ 1000 coaches numbered in increasing order 1, 2, . . . , N. The chief for train reorganizations must know whether it is possible to marshal coaches continuing in the direction B so that their order will be a1.a2, . . . , aN . Help him and write a program that decides whether it is possible to get the required order of coaches. You can assume that single coaches can be disconnected from the train before they enter the station and that they can move themselves until they are on the track in the direction B. You can also suppose that at any time there can be located as many coaches as necessary in the station. But once a coach has entered the station it cannot return to the track in the direction A and also once it has left the station in the direction B it cannot return back to the station.

Input

The input file consists of blocks of lines. Each block except the last describes one train and possibly more requirements for its reorganization. In the first line of the block there is the integer N described above. In each of the next lines of the block there is a permutation of 1, 2, . . . , N. The last line of the block contains just ‘0’. The last block consists of just one line containing ‘0’.

Output

The output file contains the lines corresponding to the lines with permutations in the input file. A line of the output file contains ‘Yes’ if it is possible to marshal the coaches in the order required on the corresponding line of the input file. Otherwise it contains ‘No’. In addition, there is one empty line after the lines corresponding to one block of the input file. There is no line in the output file corresponding to the last “null” block of the input file.

Sample Input

5

1 2 3 4 5

5 4 1 2 3

0

6

6 5 4 3 2 1

0

0

Sample Output

Yes

No

Yes

题目类型：栈模拟

算法分析：使用一个Stack数组表示序列中数的情况，Stack[i] == 0表示i元素还没有入栈；Stack[i] == 1表示i元素在栈内；Stack[i] == 2表示i元素已经出栈。边输入边处理：维护一个合法标志valid，初始时valid置为true。读入一个元素a，判断比a大的元素是否还有在栈中的，如果还有则valid置为false；如果判断完之后valid仍为true，则更新在栈中的最大值元素maxval，a出栈并将小于a的所有还没有入栈的元素都压入栈中。另一个解法是考虑每个元素进入栈中只有两种操作：(1)进栈然后出栈(2)进栈然后暂不出栈。也即存在进栈，立即出栈和暂缓出栈三种方式。依次考察出栈序列的每个元素并判断。下面第一份代码是更好的实现