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)

#pragma GCC optimize("Ofast","inline","-ffast-math")
#pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
#define int long long
using namespace __gnu_pbds;
using namespace std;

int n,a[210],num[210][70];
char c[210];

signed main()
{
	ios::sync_with_stdio(false);
	int t;
	cin>>t;
	while(t--){
		cin>>n;
		for(int i=1;i<=n;i++) cin>>a[i];
		for(int i=1;i<=n;i++) cin>>c[i];
		for(int i=1;i<=n;i++){
			int x=a[i];
			for(int j=1;j<=64;j++){
				num[i][j]=x%2;x/=2; 
			}
		}
		bool flag=true;
		for(int i=1;i<=64;i++){
			int res=-1;
			for(int j=1;j<=n;j++){
				if(num[j][i]==1) res=c[j]-'0';
			}
			if(res==1){
				flag=false;break;
			}
		}
		if(!flag) cout<<1<<'\n';
		else{
			bool flag=true;
			for(int i=1;i<=64;i++){
				for(int j=i+1;j<=64;j++){
					int res=-1;
					for(int k=1;k<=n;k++){
						if(num[k][i]!=num[k][j]) res=c[k]-'0';
					}
					if(res==1){
						flag=false;break;
					}
				}
				if(!flag) break;
			}
			if(!flag) cout<<1<<'\n';
			else cout<<0<<'\n';
		}
	}	
	return 0;
}