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: Time Limit Exceeded)

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <fstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
 
int solve()
{
	int n;
	cin >> n;

	vector<ll> a(n, 0);
	for(int i = 0; i < n; i++) {
		cin >> a[i];
	}
	string s;
	cin >> s;

	vector<ll> zero, one;
	for(int i = 0; i < n; i++) {
		if(s[i] == '0') {
			zero.push_back(a[i]);
		} else {
			one.push_back(a[i]);
		}
	}

	vector<ll> zeros, ones;
	zeros.push_back(0);
	ones.push_back(0);

	for(int i = 0; i < (int)zero.size(); i++) {
		int szero = (int)zeros.size();
		vector<ll> tmp = zeros;
		if(!binary_search(tmp.begin(), tmp.end(), zero[i])) {
			for(int j = 0; j < szero; j++) {
				ll x = zero[i] ^ zeros[j];
				if(!binary_search(tmp.begin(), tmp.end(), x)) {
					tmp.push_back(x);
					sort(tmp.begin(), tmp.end());
				}
			}
		}
		zeros = tmp;
	}

	for(int i = 0; i < (int)one.size(); i++) {
		int sone = (int)ones.size();
		vector<ll> tmp = ones;
		if(!binary_search(tmp.begin(), tmp.end(), one[i])) {
			for(int j = 0; j < sone; j++) {
				ll x = one[i] ^ ones[j];
				if(!binary_search(tmp.begin(), tmp.end(), x)) {
					tmp.push_back(x);
					sort(tmp.begin(), tmp.end());
				}
			}
		}
		ones = tmp;
	}

	int res = 0;
	for(int i = 0; i < (int)ones.size(); i++) {
		if(!binary_search(zeros.begin(), zeros.end(), ones[i])) {
			res = 1;
			break;
		}
	}
	cout << res << endl;

	return 0;
}

int main()
{
	int t;
	cin >> t;

	for(int i = 0; i < t; i++) {
		solve();
	}

    return 0;
}