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 befalse, 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.

Fig. 1: An example of FBT with the maximum depth 4 and sequential node numbers.

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:

Input

Contains l+2 lines.

Line 1 		 I the number of test cases

Line 2 		 

test case #1, two decimal numbers that are separatedby one blank

...

Line k+1 

test case #k

Line l+1 

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

题解：一个小球从二叉树的顶端开始往下落，刚开始二叉树结点开关闭合，如果闭合向左树落，否则右树，问第I个球会落到哪个结点；递归超时；

当然也可以把球数看为切入点，每次球往下落，落到此结点的次数是上一个节点次数的一半；

ac代码：

 #include<iostream>

#include<cstdio>

#include<algorithm>

#include<cstring>

#include<cmath>

#include<queue>

#include<stack>

#include<vector>

using namespace std;

const int INF=0x3f3f3f3f;

const double PI=acos(-1.0);

#define mem(x,y) memset(x,y,sizeof(x))

#define SI(x) scanf("%d",&x)

#define SL(x) scanf("%lld",&x)

#define PI(x) printf("%d",x)

#define PL(x) printf("%lld",x)

#define P_ printf(" ");

#define T_T while(T--)

int main(){

	int D,I;

	int T;

	while(SI(T),T!=-1){

		T_T{

		SI(D);SI(I);

		int cur=1,root=1;

		while(++cur<=D){

			if(I%2==1){

				root=root<<1;

				}

			else root=root<<1|1;

			I=(I+1)/2;

			}

		printf("%d\n",root);

		}

	}

	return 0;

}

　超时代码：因为球的数目可能很多，对于每一个暴力，肯定要超时了

#include<iostream>

#include<cstdio>

#include<algorithm>

#include<cstring>

#include<cmath>

#include<queue>

#include<stack>

#include<vector>

using namespace std;

const int INF=0x3f3f3f3f;

const double PI=acos(-1.0);

#define mem(x,y) memset(x,y,sizeof(x))

#define SI(x) scanf("%d",&x)

#define SL(x) scanf("%lld",&x)

#define PI(x) printf("%d",x)

#define PL(x) printf("%lld",x)

#define P_ printf(" ");

#define T_T while(T--)

#define lson root<<1

#define rson root<<1|1

typedef long long LL;

int D,I;

const int MAXN=1<<20;

int tree[MAXN];

int ans;

void query(int root,int d){

	if(d==D){

		ans=root;

		return;

	}

	if(!tree[root]){

		tree[root]=1;

		query(lson,d+1);

	}

	else{

		tree[root]=0;

		query(rson,d+1);

	}

}

int main(){

	int T;

	while(SI(T),T!=-1){

		T_T{

			mem(tree,0);

			SI(D);SI(I);

			while(I--){

				query(1,1);

			}

			printf("%d\n",ans);

		}

	}

	return 0;

}