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;

typedef double D;
typedef long long int LL;

#define st first
#define nd second
#define PLL pair <LL, LL>
#define PII pair <int, int>

const int N = 1e6 + 7;
const int MX = 1e9 + 7;
const LL INF = 1LL * MX * MX;

int n;
string s;
LL val[N];

void solve(){
	cin >> n;
	for(int i = 0; i < n; ++i)
		cin >> val[i];
	cin >> s;
	
	vector <LL> vals;
	for(int i = n - 1; i >= 0; --i){
		LL cur = val[i];
		for(auto v: vals)
			if((cur ^ v) < cur)
				cur ^= v;
		
		if(s[i] == '1'){
			if(cur){
				cout << "1\n";
				return;
			}
		}
		else if(cur){
			vals.push_back(cur);
			sort(vals.begin(), vals.end());
			reverse(vals.begin(), vals.end());
		}
	}
	
	cout << "0\n";
}

int main(){
	int cases;
	cin >> cases;
	
	while(cases--)
		solve();
	return 0;
}