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;
typedef long long int ll;
typedef long double ld;
typedef pair<ll,ll> p2;
ll basis[60];
ll insert(ll mask){
	ll p=1;
	for(ll i=0;i<60;i++){
		if(!(mask&p)){
			p*=2;
			continue;
		}
		if(!basis[i]){
			basis[i]=mask;
			return 1;
		}
		mask^=basis[i];
		p*=2;
	}
	return 0;
}
void solve(){
	ll n;
	cin>>n;
	string s;
	vector<ll> v(n);
	for(ll i=0;i<n;i++)
	cin>>v[i];
	cin>>s;
	for(ll i=n-1;i>=0;i--)
	if(s[i]=='0')
	insert(v[i]);
	else if(s[i]=='1'){
		if(!insert(v[i]))
		continue;
		cout<<"1\n";
		return;
	}
	cout<<"0\n";
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ll qc=1;
    cin>>qc;
    for(ll i=1;i<=qc;i++){
		solve();
		for(ll j=0;j<60;j++)
		basis[j]=0;
	}	
}