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;

struct base{
	long long arr[63]; base(){memset(arr , 0 , sizeof(arr));}
	bool ins(long long val){
		for(int i = 60 ; ~i ; --i)
			if(val >> i & 1)
				if(arr[i]) val ^= arr[i];
				else{arr[i] = val; return 1;}
		return 0;
	}
	bool find(long long val){
		for(int i = 60 ; ~i ; --i)
			if(val >> i & 1)
				if(arr[i]) val ^= arr[i];
				else return 0;
		return 1;
	}
};

int main(){
	int T , N;
	for(cin >> T ; T ; --T){
		cin >> N; long long A[203]; string str;
		for(int i = 0 ; i < N ; ++i) cin >> A[i];
		cin >> str; bool flg = 0 , win = 0; base now;
		for(int i = N - 1 ; ~i ; --i)
			if(str[i] == '1') win |= !now.find(A[i]);
			else flg |= !now.ins(A[i]);
		cout << win << endl;
	}
	return 0;
}