Problem: 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.

Example

Input

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


Output

1
0
0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

ll T;
ll N;
vector<ll> A;
string S;
vector<ll> M;

void join(ll a) {
  for (ll i = 0; i < M.size(); ++i) {
    if ((M[i] ^ a) < a) a ^= M[i];
  }
  if (a != 0) {
    M.push_back(a);
    sort(M.begin(), M.end(), greater<ll>());
  }
}
bool check(ll a) {
  for (ll i = 0; i < M.size(); ++i) {
    if ((M[i] ^ a) < a) a ^= M[i];
  }
  return a == 0;
}
ll solve() {
  M.clear();
  for (ll i = N - 1; i >= 0; --i) {
    if (S[i] == '0') {
      join(A[i]);
    } else {
      if (!check(A[i])) return 1;
    }
  }
  return 0;
}
int main() {
  cin >> T;
  for (ll t = 0; t < T; ++t) {
    cin >> N;
    A.resize(N);
    for (ll i = 0; i < N; ++i) cin >> A[i];
    cin >> S;
    cout << solve() << endl;
  }
  return 0;
}