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 <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <utility>
#include <queue>
#define maxn 205
using namespace std;
typedef long long ll;
typedef long double ldouble;

int n;
char s[maxn];
ll b[65], a[maxn];
bool ins(ll x) { // increase?
  for (int i = 59; i >= 0; i--) {
    if (x >> i) {
      if (!b[i]) {
        b[i] = x;
        return true;
      }
      x ^= b[i];
    }
  }
  return false;
}
void solve() {
  memset(b, 0, sizeof(b));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  scanf("%s", s + 1);
  for (int i = n; i >= 1; i--) {
    if (s[i] == '0') ins(a[i]);
    else {
      if (ins(a[i])) {
        puts("1");
        return;
      }
    }
  }
  puts("0");
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}