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 maxn = 200 + 10;
ll a[maxn];

const int L = 63;
ll d[L];
bool add(ll x){
    for(int i = L - 1; i >= 0; i--) if(x & 1ll << i){
        if(d[i] == -1){d[i] = x; return true;}
        x ^= d[i];
    }
    return false;
}

int main(){
//    Fast;
    int t; scanf("%d", &t); while(t--){
        memset(d, -1, sizeof d);

        int n; scanf("%d", &n);
        for(int i = 0; i < n; i++) scanf("%lld", a + i);
        char s[maxn]; scanf("%s", s);
        int flag = 0;
        for(int i = n - 1; i >= 0; i--){
            if(add(a[i]) && s[i] == '1'){
                flag = 1;
                break;
            }
        }
        printf("%d\n", flag);
    }
    return 0;
}