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 rep(i,n) for(int i=0; i<(n); i++)
using namespace std;
using ll = long long;

int main(){
    int t;cin >> t;
    
    rep(test,t){
        int n; cin >> n;
        vector<ll> a(n);
        rep(i,n)cin >> a[i];
        string s; cin >> s;
        
        vector<ll> dp;//遷移は基底が増えるだけ
        for(int i=n-1; i>=0; i--){
            for(ll b: dp)a[i] = min(a[i], a[i]^b);
            //minを取れば掃き出しが可能
            if(s[i]=='0'){
                dp.push_back(a[i]);//0は何個あってもいい
            }else if(a[i]!=0){
                //1のターンで新しい基底を取れたなら終了
                dp.clear();
                break;
            }
        }
        if(dp.empty())
            cout << 1 << endl;
        else
            cout << 0 << endl;
    }
    
    return 0;
}