Problem: There are two persons, numbered 0 and 1, and a variable x whose initial value is 0. The two persons now play a game. The game is played in N rounds. The following should be done in the i-th round (1 \leq i \leq N):

* Person S_i does one of the following:
* Replace x with x \oplus A_i, where \oplus represents bitwise XOR.
* Do nothing.



Person 0 aims to have x=0 at the end of the game, while Person 1 aims to have x \neq 0 at the end of the game.

Determine whether x becomes 0 at the end of the game when the two persons play optimally.

Solve T test cases for each input file.

Constraints

* 1 \leq T \leq 100
* 1 \leq N \leq 200
* 1 \leq A_i \leq 10^{18}
* S is a string of length N consisting of `0` and `1`.
* All numbers in input are integers.

Input

Input is given from Standard Input in the following format. The first line is as follows:


T


Then, T test cases follow. Each test case is given in the following format:


N
A_1 A_2 \cdots A_N
S


Output

For each test case, print a line containing `0` if x becomes 0 at the end of the game, and `1` otherwise.

Example

Input

3
2
1 2
10
2
1 1
10
6
2 3 4 5 6 7
111000


Output

1
0
0

Solution (in CPP):

Solution:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define llong long long
int T;
int n;
llong a[220];
llong B[110];
int op[220];
void ins(llong x)
{
	for(int i=61;i>=0;--i)
		if((x>>i)&1)
		{
			if(!B[i])
			{
				B[i]=x;
				break;
			}
			x^=B[i];
		}
}
bool find(llong x)
{
	for(int i=61;i>=0;--i)
		if((x>>i)&1)x^=B[i];
	return x==0;
}
int main()
{
	scanf("%d",&T);
while(T--)
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i)scanf("%lld",&a[i]);
	for(int i=1;i<=n;++i)scanf("%1d",&op[i]);
	memset(B,0,sizeof(B));
	int ans=0;
	for(int i=n;i;--i)
		if(op[i]==0)ins(a[i]);
		else if(!find(a[i]))
		{
			ans=1;break;
		}
	printf("%d\n",ans);
}
	return 0;
}