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 11.0.6) (judge verdict: Runtime Error)

import java.util.Scanner;
        
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        
        int t = sc.nextInt();
        int[] answer = new int[t];

        for (int i = 0; i < t; i++) {
            long x = 0;
            int n = sc.nextInt();
            long[] aArray = new long[n];
            for (int j = 0; j < n; j++) {
                aArray[j] = sc.nextInt();
            }
            String s = sc.next();
            String[] sArray = new String[n];
            for (int j = 0; j < n; j++) {
                sArray[j] = s.substring(j, j + 1);
            }

            for (int j = 0; j < n; j++) {
                if (sArray[j].equals("0")) {
                    // x = 0にしたい人のターン
                    if ((x ^ aArray[j]) == 0) {
                        x = (long)(x ^ aArray[j]);
                    }
                } else {
                    // x = 0にしたくない人のターン
                    if ((x ^ aArray[j]) != 0) {
                        x = (long)(x ^ aArray[j]);
                    }
                }
            }
            if (x == 0) {
                answer[i] = 0;
            } else {
                answer[i] = 1;
            }
        }
        for (int i = 0; i < t; i++) {
            System.out.println(answer[i]);
        }
    }
}