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 <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <array>

using namespace std;
using ll = long long;

int main()
{
    ll T;
    cin >> T;
    for (ll _i = 1; _i <= T; _i++)
    {
        ll N;
        array<ll, 400> A;
        string S;

        cin >> N;
        for (ll _j = 1; _j <= N; _j++) cin >> A[_j];

        cin >> S;

        set<ll> ok;
        set<ll> kisos;
        ll kiso = 0;
        bool failed = false;

        for (ll _j = N; _j >= 1; _j--)
        {
            ll a = A[_j];
            if (S[_j - 1] == '0')
            {
                vector<ll> ok2;
                for (ll k : ok) ok2.push_back(k);
                for (ll k : kisos) ok2.push_back(k);

                for (ll k : ok2)
                {
                    ll x = k ^ a;
                    if ((x & (x - 1)) == 0)
                    {
                        kiso |= x;
                        kisos.insert(x);
                    }
                    else
                    {
                        ok.insert(x);
                    }
                }

                if ((a & (a - 1)) == 0)
                {
                    kiso |= a;
                    kisos.insert(a);
                }
                else
                {
                    ok.insert(a);
                }

                ok2.clear();
                for (ll k : ok) ok2.push_back(k);

                for (ll k : ok2)
                {
                    if ((~kiso & k) == 0) ok.erase(k);
                    for (ll kk : ok2)
                    {
                        if (ok.count(kk) && ok.count(k ^ kk)) ok.erase(k);
                    }
                }

                //cerr << kiso << endl;
                //for (ll k : ok) cerr << k << ", ";
                //cerr << endl;
            }
            else if (S[_j - 1] == '1')
            {
                if (((~kiso & a) != 0) && ok.count(a) == 0)
                {
                    bool f2 = false;
                    for (ll k : ok) if (ok.count(a ^ k)) {
                        f2 = true;
                        break;
                    }
                    if (!f2)
                    {
                        failed = true;
                        break;
                    }
                }
            }
        }

        cout << (failed ? 1 : 0) << endl;
    }
}