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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef vector<int> vi;
typedef vector<vector<int> > vvi;
typedef vector<pair<int, int> > vpii;
typedef pair<int, int> pii;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef vector<ll> vll;
#define INF 0x3f3f3f3f
#define MOD 1000000007LL
#define EPS 0.00001
#define f first
#define s second
#define pb push_back

#define FOR(i, a, b) for (int i=(a); i<=(signed)(b); i++)
#define F0R(i, a) for (int i=0; i<(signed)(a); i++)
#define RFOR(i, a, b) for (int i=(a); i >= b; i--)

#define MN 205
int t, n;
ll a[MN];
string s;
vll basis;

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    cin >> t;
    F0R(_, t){
        basis.clear();
        cin >> n;
        F0R(i, n){
            cin >> a[i];
        }
        cin >> s;
        bool ok = true;
        RFOR(i, n-1, 0){
            if(s[i] == '0'){
                ll A = a[i];
                for(ll b : basis){
                    A = min(A, A^b);
                }
                if(A){
                    int ind = 0;
                    while(ind < basis.size() && basis[ind] > A) ++ind;
                    basis.insert(basis.begin()+ind, A);
                }
            } else{
                ll A = a[i];
                for(ll b : basis){
                    A = min(A, A^b);
                }
                if(A){
                    ok = false;
                    break;
                }
            }
        }
        if(ok){
            cout << "0\n";
        } else{
            cout << "1\n";
        }
    }

    return 0;
}