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)

#include <bits/stdc++.h>

#define F first
#define S second
#define MP make_pair
#define pb push_back
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define LCM(a, b) (a) / __gcd((a), (b)) * (b)
#define CEIL(a, b) (a)/(b)+(((a)%(b))?1:0)
#define ln '\n'

using namespace std;
using LL = long long;
using ldouble = long double;
using P = pair<int, int>;
using LP = pair<LL, LL>;

static const int INF = INT_MAX;
static const LL LINF = LLONG_MAX;
static const int MIN = INT_MIN;
static const LL LMIN = LLONG_MIN;
static const int MOD = 1e9 + 7;
static const int SIZE = 200005;

const int dx[] = {0, -1, 1, 0};
const int dy[] = {-1, 0, 0, 1};

vector<LL> Div(LL n) {
    vector<LL> ret;
    for(LL i = 1; i * i <= n; ++i) {
        if(n % i == 0) {
            ret.pb(i);
            if(i * i != n) ret.pb(n / i);
        }
    }
    sort(all(ret));
    return ret;
}

void solve() {
    int N;
    cin >> N;
    vector<LL> v(N);
    for(int i = 0; i < N; ++i) {
        cin >> v[i];
    }

    string s;
    cin >> s;

    int p0[100] = {}, p1[100] = {};
    for(int i = N - 1; i >= 0; --i) {
        LL t = v[i];
        if(s[i] == '0') {
            int b = 0;
            while(t > 0) {
                if(t & 1) p0[b]++;
                t >>= 1;
                ++b;
            }
        } else {
            int b = 0;
            while(t > 0) {
                if(t & 1) {
                    p1[b]++;
                    if(p0[b] == 0) {
                        cout << 1 << endl;
                        return;
                    }
                }
                t >>= 1;
                ++b;
            }
        }
    }
    cout << 0 << endl;

}

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

    int t;
    cin >> t;
    while(t--) {
        solve();
    }

    return 0;
}