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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
using namespace std;

#define ALL(v) begin(v),end(v)

typedef long long LL;
typedef vector<LL> vll;

int solve(){
	int n;
	cin >> n;
	vll as(n);
	for(int i = 0; i < n; ++i){
		cin >> as[i];
	}
	string s;
	cin >> s;
	vll bs;
	for(int i = n - 1; i >= 0; --i){
		LL a = as[i];
		for(LL x : bs){
			if((a ^ x) < a){ a ^= x; }
		}
		if(s[i] == '0'){
			if(a != 0){
				bs.insert(lower_bound(ALL(bs), a, greater<LL>()), a);
			}
		}
		else{
			if(a != 0){ return 1; }
		}
	}
	return 0;
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	int t;
	cin >> t;
	for(int i = 0; i < t; ++i){
		cout << solve() << '\n';
	}
}