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 C (Clang 10.0.0) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
int N;
unsigned long long A[201];
unsigned int continuous = 0;

int process(unsigned long long a, int con) {

	if (a == 0) {
		return 2;
	}
	if (con >= continuous) {
		return 0;
	}

	unsigned long long ab = a ^ A[N - con-1];
	//printf("-%llu/", ab);
	int n = con + 1;

	if (2 == process(ab, n)) {
		return 2;
	}

	if (2 == process(a, n)) {
		return 2;
	}

	return 0;
}

int main(void) {
	int T;

	//unsigned long long A[200];
	char S[201];
	unsigned long long x = 0;
	int result[200] = { 0 };

	scanf("%d", &T);

	for (int i = 0; i < T; i++) {
		scanf("%d", &N);

		for (int j = 0; j < N; j++) {
			scanf("%llu", &A[j]);
		}

		scanf("%s", S);

		if (S[N-1] == '0') {
			continuous = 1;
			for (int j = 1; S[N - 1 - j] == '0'; continuous++, j++) {

			}
		}
		else {
			continuous = 0;
		}

		for (int j = 0; j < N - continuous; j++) {
			if (S[j] == '0') {//target x=0
				if (x == 0) {

				}
				else {
					x = x ^ A[j];
				}
			}
			else {//target x=1
				if (x != 0) {

				}
				else {
					x = x ^ A[j];
				}
			}
				//printf("x:%llu a:%llu s:%c\n", x,A[j],S[j]);
		}

		if (process(x, 0) == 2) {
			x = 0;
		}
		else {
			x = 1;
		}

		if (x == 0) {
			result[i] = 0;
		}
		else {
			result[i] = 1;
		}
		x = 0;
		continuous = 0;
	}

	for (int i = 0; i < T; i++) {
		printf("%d\n", result[i]);
	}

	return 0;
}