Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

3
2
1 2
10
2
1 1
10
6
2 3 4 5 6 7
111000

Sample Output 1

1
0
0

In the first test case, if Person 1 replaces x with 0 \oplus 1=1, we surely have x \neq 0 at the end of the game, regardless of the choice of Person 0.

In the second test case, regardless of the choice of Person 1, Person 0 can make x=0 with a suitable choice.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

struct base{
	long long arr[63]; base(){memset(arr , 0 , sizeof(arr));}
	bool ins(long long val){
		for(int i = 60 ; ~i ; --i)
			if(val >> i & 1)
				if(arr[i]) val ^= arr[i];
				else{arr[i] = val; return 1;}
		return 0;
	}
	bool find(long long val){
		for(int i = 60 ; ~i ; --i)
			if(val >> i & 1)
				if(arr[i]) val ^= arr[i];
				else return 0;
		return 1;
	}
};

int main(){
	int T , N;
	for(cin >> T ; T ; --T){
		cin >> N; long long A[203]; string str;
		for(int i = 0 ; i < N ; ++i) cin >> A[i];
		cin >> str; bool flg = 0 , win = 0; base now;
		for(int i = N - 1 ; ~i ; --i)
			if(str[i] == '1') win |= !now.find(A[i]);
			else flg |= !now.ins(A[i]);
		cout << win << endl;
	}
	return 0;
}