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;

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	int tc;
	cin >> tc;
	while(tc--){
		int n;
		cin >> n;
		vector<long long> a(n);
		for(int i=0;i<n;i++) cin >> a[i];
		string str;
		cin >> str;
		
		if(str.back()=='1'){
			cout << 1 << "\n";
			continue;
		}
		
		vector<long long> base;
		for(int i=n-1;i>=0;i--){
			long long v=a[i];
			for(long long b:base){
				v=min(v,v^b);
			}
			if(str[i]=='0' && v>0){
				for(int j=0;j<(int)base.size();j++){
					base[j]=min(base[j],base[j]^v);
				}
				base.push_back(v);
				sort(base.begin(),base.end());
				reverse(base.begin(),base.end());
			}
			a[i]=v;
		}
		
		bool bo=false;
		for(int i=0;i<n;i++){
			if(str[i]=='1' && a[i]!=0) bo=true;
		}
		
		if(bo) cout << 1 << "\n";
		else cout << 0 << "\n";
	}
	return 0;
}