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>
typedef long long LL;
using namespace std;

const int N=210;
int n;
LL a[N];
char str[N];
struct Base {
	LL arr[62];
	void init() { memset(arr,0,sizeof(arr)); }
	void ins(LL x) {
		for(int i=61;~i;--i)
			if(x>>i&1) {
				if(!arr[i]) { arr[i] = x; break; }
				else x ^= arr[i];
			}
	}
	bool find(LL x) {
		for(int i=61;~i;--i)
			if(x>>i&1) x ^= arr[i];
		return x == 0;
	}
} A;
int main() {
	int T; scanf("%d",&T);
	while(T--) {
		scanf("%d",&n);
		for(int i=1;i<=n;++i)
			scanf("%lld",&a[i]);
		scanf("%s",str+1);
		A.init();
		bool ans=0;
		for(int i=n;i>=1;--i) {
			if(str[i] == '0') A.ins(a[i]);
			else ans |= !A.find(a[i]);
		}
		cout<<ans<<endl;
	}
	return 0;
}