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>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

int matrix_rank(vector<unsigned long long> A){
	int n=A.size(),r=0;
	rep(i,64){
		int piv=-1;
		for(int j=r;j<n;j++) if(A[j]>>i&1) {
			piv=j;
			break;
		}
		if(piv==-1) continue;

		for(int j=r;j<n;j++) if(j!=piv && A[j]>>i&1) {
			A[j]^=A[piv];
		}
		swap(A[r],A[piv]);
		r++;
	}
	return r;
}

void solve(){
	int n; cin>>n;
	vector<unsigned long long> a(n);
	rep(i,n) cin>>a[i];
	string s; cin>>s;

	vector<unsigned long long> A;
	int r=0;
	for(int i=n-1;i>=0;i--){
		A.emplace_back(a[i]);
		int r2=matrix_rank(A);
		if(r<r2 && s[i]=='1'){
			puts("1");
			return;
		}
		r=r2;
	}
	puts("0");
}

int main(){
	int q; scanf("%d",&q); rep(_,q) solve();
	return 0;
}