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 ll long long
using namespace std;

void solve() {
	int n;
	cin >> n;
	ll a[n], base[65];
	memset(base, 0, sizeof base);
	for(int i = 0; i < n; ++i)cin >> a[i];
	string s; cin >> s;
	for(int i = n - 1; ~i; --i) {
		bool contains = true;
		for(int bit = 0; bit < 61; ++bit) {
			if(a[i] >> bit & 1) {
				if(!base[bit]) {
					contains = false;
					base[bit] = a[i];
					break;
				}
				a[i] ^= base[bit];
			}
		}
		if(!contains) {
			if(s[i] == '1') {
				cout << "1\n";
				return;
			}
		}
	}
	cout << "0\n";
}

int main() {
	int t; cin >> t;
	while(t--) {
		solve();
	}
}