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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(){
    int t;
    cin>>t;
    while(t--){
        int n;
        ll x=0;
        cin>>n;
        ll a[n];
        for(int i=0;i<n;++i)cin>>a[i];
        string s;
        cin>>s;
        unordered_set<ll> S;
        for(int i=0;i<n;++i){
            if(s[i]=='0')S.insert(a[i]);
        }
        
        
        for(int i=0;i<n;++i){
            if(s[i]=='1'){
                if(x==0) x=(x^a[i]);
                else {
                    if(S.find(x)!=S.end()) x=(x^a[i]);
                    else {;}
                }
            }
            if(s[i]=='0'){
                if(x==0){S.erase(S.find(a[i]));continue;}
                if(x==a[i]) x=(x^a[i]);
                else {
                    if(S.find(x)!=S.end()){;}
                    else if(S.find(x^a[i])!=S.end()) x=(x^a[i]);
                    else {;}
                }
                S.erase(S.find(a[i]));
            }
        }
        if(x==0)cout<<0<<endl;
        else cout<<1<<endl;
    }
    return 0;
}