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;

typedef long long LL;
#define N 300

const LL k=60;
LL n,a[N],f[80];
char s[N];

void ins(LL x){
	for (LL i=k;~i;--i){
		if (x>>i&1){
			if (!f[i]){f[i]=x; return;}
			else x^=f[i];
		}
	}
}

bool pd(LL x){
	for (LL i=k;~i;--i){
		if (x>>i&1){
			if (!f[i]) return 0;
			x^=f[i];
		}
	}
	return 1;
}

void solve(){
	scanf("%lld",&n);
	for (LL i=1;i<=n;++i) scanf("%lld",a+i);
	scanf("%s",s+1);
	memset(f,0,sizeof f);
	LL flag=0;
	for (LL i=n;i;--i){
		if (s[i]=='0') ins(a[i]);
		else if (!pd(a[i])){flag=1; break;}
	}
	printf("%lld\n",flag);
}

int main(){
	LL T; scanf("%lld",&T);
	while (T--) solve();
	
	return 0;
}