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;

const int maxn = 100010;
typedef long long ll;

int T, n;
ll a[maxn];
char s[maxn];

ll b[61];

void add(ll x) {
	for (int i = 60; i >= 0; i--) {
		if ((x >> i) & 1) {
			if (!b[i]) {
				b[i] = x;
				break;
			} else x ^= b[i];
		}
	}
}

int check(ll x) {
	for (int i = 60; i >= 0; i--) {
		if ((x >> i) & 1) {
			if (!b[i]) return 0;
			else x ^= b[i];
		}
	}
	return 1;
}

int main() {
	scanf("%d", &T);
	while (T--) {
		scanf("%d", &n);
		for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
		scanf("%s", s+1);
		for (int i = 0; i <= 60; i++) b[i] = 0;
		int ok = 1;
		for (int i = n; i >= 1; i--) {
			if (s[i] == '0') {
				add(a[i]);
			} else if (!check(a[i])) {
				ok = 0;
				break;
			}
		}
		printf("%d\n", 1-ok);
	}
	return 0;
}