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
#include <bits/stdc++.h>
using namespace std;
typedef long long i64;
const int MAX = 1e6 + 10;
const int MOD = 1e9 + 7;
const int oo = 1e9;

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	#define endl '\n'
	int T; cin >> T;
	while(T--){
		int n; cin >> n;
		vector<i64> A(n);
		for(int i = 0; i <n; i++)cin >> A[i];
		string p;
		cin >> p;
		vector<i64> basis;
		int ans = 0;
		for(int i = n - 1; i >= 0; i--){
			i64 c = A[i];
			for(auto b: basis) c= min(c, c ^ b);

			if(p[i] == '1'){
				if(c > 0){
					ans = 1;break;
				}
			}
			else{
				if(c > 0)basis.push_back(c);
			}
		}
		cout << ans << endl;

		
	}
}
