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;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
using Int = long long;
const char newl = '\n';

//INSERT ABOVE HERE
signed solve(){
  Int n;
  cin>>n;
  vector<Int> as(n);
  for(Int i=0;i<n;i++) cin>>as[i];

  string s;
  cin>>s;

  vector<Int> bs;
  for(Int i=n-1;i>=0;i--){
    Int a=as[i];
    for(Int b:bs)
      chmin(a,a^b);

    if(s[i]=='0'){
      if(a!=0) bs.emplace_back(a);
    }

    if(s[i]=='1'){
      if(a!=0){
        cout<<'1'<<newl;
        return 0;
      }
    }
  }
  cout<<'0'<<newl;
  return 0;
}

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

  Int T;
  cin>>T;
  while(T--) solve();

  return 0;
}