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)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
signed main() { 
  ios::sync_with_stdio(false); cin.tie(0);
  auto solve = [] () {
    int n;
    cin >> n;
    vector<long long> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    string s;
    cin >> s;
    vector<long long> base(61);
    for (int i = n - 1; i >= 0; i--) {
      int cur = a[i];
      for (int j = 0; j < 61; j++) {
        if (cur & (1LL << j)) {
          cur ^= base[j];
        }
      }
      if (s[i] == '0') {
        if (cur != 0) {
          int id = 0;
          while (true) {
            if (cur & (1LL << id)) {
              break;
            }
            id++;
          }
          base[id] = cur;
        }
      } else if (cur != 0) {
        return 1;
      }
    }
    return 0;
  };
  int t;
  cin >> t;
  while (t--) cout << solve() << '\n';
  return 0;
}