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 <iostream>
#include <string>
#include <vector>
using namespace std;
typedef long long LL;

template<typename T>
istream& operator >> (istream& is, vector<T>& vs){
  for(T& v: vs) is >> v;
  return is;
}

LL extract_base(LL v, const vector<LL>& base) {
  for (LL b : base) {
    v = min(v, v^b);
  }
  return v;
}

int solve() {
  int N; cin >> N;
  vector<LL> A(N); cin >> A;
  string S; cin >> S;

  vector<LL> base;
  for (int i = N - 1; i >= 0; --i) {
    LL a = A[i];
    if (S[i] == '0') {
      a = extract_base(a, base);
      if (a != 0) base.push_back(a);
    } else {
      a = extract_base(a, base);
      if (a != 0) return 1;
    }
  }
  return 0;
}

int main() {
  int N; cin >> N;
  for (int i = 0; i < N; ++i) {
    cout << solve() << endl;;
  }
}