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: Accepted)

import java.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(br.readLine());
		label:
		for (int x = 0; x < t; x++) {
			int n = Integer.parseInt(br.readLine());
			String[] sa = br.readLine().split(" ");
			long[] a = new long[n];
			for (int i = 0; i < n; i++) {
				a[i] = Long.parseLong(sa[i]);
			}
			char[] s = br.readLine().toCharArray();

			for (int i = n - 1; i >= 0; i--) {
				if (s[i] == '0') {
					for (int k = 0; k < 61; k++) {
						if ((a[i] >> k & 1) == 1) {
							for (int j = 0; j < n; j++) {
								if (i != j && (a[j] >> k & 1) == 1) {
									a[j] ^= a[i];
								}
							}
						}
					}
				} else {
					if (a[i] != 0) {
						System.out.println(1);
						continue label;
					}
				}
			}
			System.out.println(0);
		}
	}
}