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<cstdio>
#include<cstring>
using namespace std;
const int maxn=205;
int T,n;char s[maxn];
long long a[maxn],b[70];
inline void add(long long x){
	for (int i=60;~i;i--) if((x>>i)&1)
	if(!b[i]){b[i]=x;break;}else x^=b[i];
}
inline bool query(long long x){
	for (int i=60;~i;i--) if((x>>i)&1) x^=b[i];
	return !x;
}
inline bool solve(){
	for (int i=n;i;i--)
	if(s[i]=='0') add(a[i]);
	else if(!query(a[i])) return 0;
	return 1;
}
int main(){
	for (scanf("%d",&T);T--;){
		memset(b,0,sizeof b);
		scanf("%d",&n);
		for (int i=1;i<=n;i++) scanf("%lld",&a[i]);
		scanf("%s",s+1);
		puts(solve()?"0":"1");
	}
	return 0;
}