Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

3
2
1 2
10
2
1 1
10
6
2 3 4 5 6 7
111000

Sample Output 1

1
0
0

In the first test case, if Person 1 replaces x with 0 \oplus 1=1, we surely have x \neq 0 at the end of the game, regardless of the choice of Person 0.

In the second test case, regardless of the choice of Person 1, Person 0 can make x=0 with a suitable choice.

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<iostream>
using namespace std;
int t, n, i, ii, ok, jj;
long long v[205], d[70][2], x;
char s[205];
int main(){
    cin>> t;
    for(; t; t--){
        cin>> n;
        for(i = 1; i <= n; i++){
            cin>> v[i];
        }
        cin>> s + 1;
        if(s[n] == '1'){
            cout<< 1;
            continue;
        }
        for(i = n; i >= 1; i--){
            x = v[i];
            for(ii = 59; ii >= 0; ii--){
                if( ( (x >> ii) & 1) == 1){
                    if(d[ii][ s[i] - '0' ] != 0){
                        x ^= d[ii][ s[i] - '0' ];
                    }
                    else{
                        d[ii][ s[i] - '0' ] = x;
                        break;
                    }
                }
            }
            ok = 0;
            for(ii = 0; ii < 59; ii++){
                x = d[ii][1];
                for(jj = 59; jj >= 0; jj--){
                    if(( (x >> jj) & 1) == 1){
                        x ^= d[jj][0];
                    }
                }
                if(x != 0){
                    ok = 1;
                    break;
                }
            }
            if(ok == 1){
                cout<<"1\n";
                break;
            }
        }
        if(i == 0){
            cout<<"0\n";
        }
    }
}