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 long long ll;
const int N = 200 + 2;
ll p[70];
bool chk(ll x) {
    for(int i = 61; i >= 0; i--)
    {
        if(!(x >> (ll)i))
            continue;
        if(!p[i]) {
            p[i] = x;
            return 1;
        }
        x ^= p[i];
    }
    return 0;
}
void Get_LB(ll x) {
    for(int i = 61; i >= 0; i--)
    {
        if(!(x >> (ll)i))
            continue;
        if(!p[i])
        {
            p[i] = x;
            break;
        }
        x ^= p[i];
    }
}

int T;
int n; ll a[N]; char s[N];

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);
		memset(p, 0, sizeof(p));
		bool ok = 1;
		for (int i = n; i >= 1; i --) {
			if (s[i] == '1') {
				if (chk(a[i])) ok = 0;
			} else {
				Get_LB(a[i]);
			}
		}
		printf("%s\n", ok ? "0" : "1");
	}
}