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: Time Limit Exceeded)

import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int T = sc.nextInt();
		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();
			Set<Long> setZero = new HashSet<Long>();
			setZero.add(0l);
			Set<Long> setOne = new HashSet<Long>();
			setOne.add(0l);
			boolean b = false;
			for (int j = N - 1; j >= 0; j--) {
				if (S.charAt(j) == '0') {
					Long[] temp = setZero.toArray(new Long[setZero.size()]);
					for (int k = 0; k < temp.length; k++) {
						long tempLong = A[j] ^ temp[k];
						setZero.add(tempLong);
					}
				} else {
					Long[] temp = setOne.toArray(new Long[setOne.size()]);
					for (int k = 0; k < temp.length; k++) {
						long tempLong = A[j] ^ temp[k];
						if (setZero.contains(tempLong) == false) {
							b = true;
							break;
						} else {
							setOne.add(tempLong);
						}
					}
				}
			}
			if (b == false) {
				System.out.println(0);
			} else {
				System.out.println(1);
			}
		}
	}
}