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;

using ll = long long;

void solve(){
	int n;
	cin >> n;
	vector<ll> a(n);
	for(int i = 0; i < n; i++){
		cin >> a[i];
	}
	string s;
	cin >> s;
	const ll B = 62;
	vector<ll> what(62);
	for(int i = n-1; i >= 0; i--){
		ll cur = a[i];
		for(int j = B-1; j >= 0; j--){
			if(what[j] && (cur & (1ll << j))){
				cur ^= what[j];
			}
		}
		if(s[i] == '0'){
			for(int j = B-1; j >= 0; j--){
				if((cur & (1ll << j))){
					what[j] = cur;
					break;
				}
			}
		} else if(s[i] == '1'){
			if(cur != 0){
				cout << "1" << '\n';
				return;
			}
		}
	}
	cout << "0" << '\n';
	return;
}

int main(){
	ios_base::sync_with_stdio(false), cin.tie(nullptr);
	int T;
	cin >> T;
	while(T--) solve();
}