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: Accepted)

#include <bits/stdc++.h>
using namespace std;
long long a[205];
char s[205];
long long dp[65];
int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        memset(dp, -1, sizeof(dp));
        int n;
        scanf("%d", &n);
        for(int i=1;i<=n;i++) scanf("%lld", &a[i]);
        scanf("%s", s+1);
        bool ans = 0;
        for(int i=n;i>=1;i--){
            if(s[i] == '0'){
                for(int j=60;j>=0;j--){
                    if((a[i]>>j)&1){
                        if(dp[j] == -1){
                            dp[j] = a[i];
                            break;
                        }
                        else a[i] ^= dp[j];
                    }
                }
            }
            else{
                long long sum = 0;
                for(int j=60;j>=0;j--){
                    if(((a[i]>>j)&1) != ((sum>>j)&1)){
                        if(dp[j] == -1) break;
                        sum ^= dp[j];
                    }
                }
                if(sum != a[i]){
                    ans = 1;
                    break;
                }
            }

        }
        printf("%d\n", ans);
    }
}