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>
#define be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007, INF=(1LL<<60);
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;
void solve(){
	ll n;
	cin >> n;
	vector<ll> a(n), dp(61, 0);
	for(int i=0;i<n;i++) cin >> a[i];
	string s;
    cin >> s;
    for(int i=n-1;i>=0;i--){
    	for(int j=60;j>=0;j--) if(a[i] >> j & 1){
    	    if(s[i] == '1' || dp[j] >> j & 1) a[i] ^= dp[j];
    	    else dp[j] = a[i], j = -1;
        }
        if(s[i] == '1' && a[i]){
        	cout << 1 << endl;
        	return;
        }
    }
    cout << 0 << endl;
    return;
}
int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while(t--) solve();
    return 0;
}