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 Java (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        int[] result = new int[t];
        for (int i=0; i<t; i++) {
            int n = sc.nextInt();
            long[] a = new long[n];
            for (int j=0; j<n; j++) {
                a[j] = sc.nextLong();
            }
            String s = sc.next();
            int count1 = 0;
            for (char b : s.toCharArray()) {
                if (b == '1') {
                    count1++;
                }
            }
            long[] x0 = new long[n-count1+1];
            int n0 = 0;
            for (int j=0; j<n; j++) {
                char sj = s.charAt(j);
                if (sj == '0') {
                    x0[n0] = a[j];
                    n0++;
                }
            }
            long[] x1 = new long[count1+1];
            x1[0] = 0;
            int n1 = 1;
            for (int j=0; j<n; j++) {
                boolean noHit = true;
                char sj = s.charAt(j);
                if (sj == '1') {
                    for (int k=0; k<n1; k++) {
                        x1[k] = x1[k] ^ a[j];
                        noHit = true;
                        Iterator<Long> iterator = new PatternItr(x0);
                        while (iterator.hasNext()) {
                            long y = iterator.next();
                            if (x1[k] == y) {
                                noHit = false;
                                break;
                            }
                        }
                        if (noHit) {
                            result[i] = 1;
                            break;
                        }
                    }
                    if (noHit) {
                        break;
                    }
                    x1[n1] = a[j];
                    noHit = true;
                    Iterator<Long> iterator = new PatternItr(x0);
                    while (iterator.hasNext()) {
                        long y = iterator.next();
                        if (x1[n1] == y) {
                            noHit = false;
                            break;
                        }
                    }
                    if (noHit) {
                        result[i] = 1;
                        break;
                    }
                    n1++;
                }
            }
        }
        for (int r : result) {
            System.out.println(r);
        }
    }

    public static class PatternItr implements Iterator<Long> {
        long[] x;
        public PatternItr(long[] x) {
            this.x = x;
            i = 0;
            max = (int)Math.pow(2, x.length);
        }
        int i;
        int max;

        @Override
        public boolean hasNext() {
            return i < max;
        }

        @Override
        public Long next() {
            i++;
            String s = Long.toBinaryString(i);
            StringBuilder builder = new StringBuilder();
            for (int j=0; j<x.length-s.length(); j++) {
                builder.append("0");
            }
            builder.append(s);
            s = builder.toString();
            long y = 0;
            for (int j=0; j<x.length; j++) {
                if (s.charAt(j) == '1') {
                    y = y ^ x[j];
                }
            }
            return y;
        }
    }
}