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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;

int main(void){
    int T, N;
    long long Abuf;
    vector<long long> A;
    string S;
    cin >> T;
    for(int itestcase=0; itestcase<T; itestcase++){
        cin >> N;
        A.clear();
        for(int i=0; i<N; i++){
            cin >> Abuf;
            A.push_back(Abuf);
        }
        cin >> S;
        vector<long long> DP;
        DP.push_back(0);
        for(int i=N-1; i>=0; i--){
            Abuf = A[i];
            for(long long v: DP)Abuf = min(Abuf, Abuf^v);
            if(S[i]=='0'){
                DP.push_back(Abuf);
            }else if(Abuf != 0LL){
                DP.clear();
                break;
            }
        }
        if(DP.empty()){
            cout << 1 << endl;
        }else{
            cout << 0 << endl;
            DP.clear();
        }
    } 
    return 0;
}