Problem: 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.

Example

Input

3
2
1 2
10
2
1 1
10
6
2 3 4 5 6 7
111000


Output

1
0
0

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
void test_case() {
	int n;
	scanf("%d", &n);
	vector<long long> a(n);
	for(int i = 0; i < n; ++i) {
		scanf("%lld", &a[i]);
	}
	char s[n+1];
	scanf("%s", s);
	assert((int) strlen(s) == n);
	vector<long long> base;
	for(int i = n - 1; i >= 0; --i) {
		// if(a[i] == 0) {
			// continue;
		// }
		long long x = a[i];
		for(long long b : base) {
			x = min(x, x^b);
		}
		if(x == 0) {
			continue;
		}
		if(s[i] == '0') {
			base.push_back(x);
		}
		else {
			puts("1");
			return;
		}
	}
	puts("0");
}

int main() {
	int T;
	scanf("%d", &T);
	while(T--) {
		test_case();
	}
}