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: Runtime Error)

#include <stdio.h>

int main(void)
{
    int T;
    int N;
    long long A[200];
    char S[201];
    char c;
    int i, j, k;
    
    int index;
    long long DP[20200];
    int record[100];
    
    
    scanf("%d%c", &T, &c);
    for(i=0; i<T; i++)
    {
        scanf("%d%c", &N, &c);
        for(j=0; j<N; j++)
            scanf("%lld%c", &A[j], &c);
        scanf("%s", S);
        
        index = 0;
        DP[index++] = 0;
        record[i] = 0;
        
        for(j=N-1; j>=0; j--)
        {
            if(S[j] == '0')
            {
                for(k=index-1; k>=0; k--)
                {
                    DP[index++] = DP[k] ^ A[j];
                }
            }
            else
            {
                for(k=0; k<index; k++)
                {
                    if(DP[k] == A[j])
                        break;
                }
                
                if(k == index)
                {
                    record[i] = 1;
                    break;
                }
            }
        }
    }
    
    
    for(i=0; i<T; i++)
        printf("%d\n", record[i]);
    
    /**/system("pause");
    
    return 0;
}