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<bits/stdc++.h>
using namespace std;

typedef long long LL;
const int N = 205, K = 62;
LL a[N];

LL rep[K];

void add(LL x) {
    for (int i=0; i<K; i++) {
        if (rep[i]) {
            if (x&(1LL<<i)) x^=rep[i];
        }
        else {
            rep[i] = x;
            break;
        }
    }
}

bool check(LL x) {
    for (int i=0; i<K; i++) {
        if ((x&(1LL<<i)) == 0)  continue;
        if (rep[i]) x^=rep[i];
        else        return false;
    }
    return true;
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    int t;
    cin>>t;

    while (t--) {
        int n;
        cin>>n;

        memset(rep, 0, sizeof rep);

        for (int i=0; i<n; i++)    cin>>a[i];
        string s;
        cin>>s;

        bool ans = 0;
        for (int i=n-1; i>=0; i--) {
            if (s[i] == '0')    add(a[i]);
            else {
                if (!check(a[i])) {
                    ans = 1;
                    break;
                }
            }
        }
        cout<<ans<<endl;
    }
}