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: Wrong Answer)

import java.io.*;
import java.math.*;
import java.util.*;
public class Main {

	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int T = input.nextInt();
		for (int i = 1; i <= T; i++) {
			int N = input.nextInt();
			long[] arr = new long[N];
			long x = 0;
			for (int j = 0 ; j < N; j++) {
				arr[j] = input.nextLong();
			}
			String S = input.next();
			for (int turn = 0; turn < N; turn++) {
				if (S.charAt(turn)=='0') {
					if (x==arr[turn]) x^=arr[turn];
					boolean found = false;
					long temp = x^arr[turn];
					for (int next = turn+1; next < N; next++) {
						if (temp==arr[next]) {
							found=true;
							break;
						}
					}
					if (found) x^=arr[turn];
				}else {
					boolean found = false;
					long temp = x^arr[turn];
					for (int next = turn+1; next < N; next++) {
						if (temp==arr[next]) {
							found=true;
							break;
						}
					}
					if (!found) x^=arr[turn]; //Kills 0's chances of making 0
				}
			}
			System.out.println(x==0?"0":"1");
		}
	}
}