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;
#define I inline int
#define V inline void
#define ll long long int
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define ROF(i,a,b) for(int i=a;i>=b;i--)
const int N=2e2+5;
struct bas{
	ll a[64];
	V init(){FOR(i,0,63)a[i]=0;}
	V ins(ll x){
		ROF(i,63,0)if(x>>i&1){
			if(!a[i])return void(a[i]=x);
			else x^=a[i];
		}
	}
	I find(ll x){
		ROF(i,63,0)if(x>>i&1){
			if(a[i])x^=a[i];
			else return false;
		}
		return true;
	}
}now;
ll a[N];
char st[N];
int T,n,ans;
int main(){
	for(scanf("%d",&T);T--;cout<<ans<<'\n'){
		scanf("%d",&n),now.init(),ans=0;
		FOR(i,1,n)scanf("%lld",a+i);
		scanf("%s",st+1);
		ROF(i,n,1)
			if(st[i]=='0')now.ins(a[i]);
			else if(!now.find(a[i]))ans=1;
	}
	return 0;
}