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<bits/stdc++.h>
using namespace std;
using ll=long long;
char s[210];
int T,n;ll a[210],bs0[64],bs1[64];
void ins(ll*b,ll x){
	for(int k=59;~k;k--)if(x>>k&1)
		if(!b[k]){b[k]=x;return;}
		else x^=b[k];
}
int chk(ll*b,ll x){
	for(int k=59;~k;k--)if((x>>k&1) && b[k])
		x^=b[k];
	return x==0;
}
int main(){
	scanf("%d",&T);
	while(T--){
		memset(bs0,0,sizeof bs0);
		memset(bs1,0,sizeof bs1);
		scanf("%d",&n);
		for(int i=1;i<=n;i++)scanf("%lld",a+i);
		scanf(" %s",s+1);
		for(int i=n;i;){
		//	memset(bs1,0,sizeof bs1);
			while(i && s[i]=='0')ins(bs0,a[i--]);
			while(i && s[i]=='1')ins(bs1,a[i--]);
			for(int i=0;i<60;i++)if(bs1[i])
				if(!chk(bs0,bs1[i])){puts("1");goto sc;}
		}
		puts("0");sc:;
	}
	return 0;
}