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>
#define RI register int
using namespace std;
typedef long long LL;
const int N=205;
struct Linear_Basis
{
	LL p[63];
	inline void clear(void)
	{
		for (RI i=0;i<63;++i) p[i]=0;
	}
	inline void insert(LL x)
	{
		for (RI i=62;~i;--i) if ((x>>i)&1)
		{
			if (!p[i]) return (void)(p[i]=x); x^=p[i];
		}
	}
	inline bool exist(LL x)
	{
		for (RI i=62;~i;--i) if ((x^p[i])<x) x^=p[i]; return !x;
	}
}l;
int t,n; LL a[N]; char s[N];
int main()
{
	for (scanf("%d",&t);t;--t)
	{
		RI i; for (scanf("%d",&n),i=1;i<=n;++i) scanf("%lld",&a[i]);
		scanf("%s",s+1); if (s[n]=='1') { puts("1"); continue; }
		bool flag=0; for (l.clear(),i=n;i;--i) if (s[i]=='1')
		{
			if (!l.exist(a[i])) { flag=1; break; }
		} else l.insert(a[i]);
		puts(flag?"1":"0");
	}
	return 0;
}