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;
typedef long long ll;
const int N=205,up=60;
int n;
ll a[N],p[N];
char s[N];
void add(ll x)
{
    for(int i=up;i>=0;i--)
        if(x>>i&1)
        {
            if(!p[i]) {p[i]=x;break;}
            else x=x^p[i];
        }
}
bool query(ll x)
{
    for(int i=up;i>=0;i--)
        if(x>>i&1)
    {
        if(!p[i]) return true;
        else x=x^p[i];
    }
    return false;
}
int main()
{
    int t;scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
        scanf("%s",s+1);
        if(s[n]=='1')
        {
            printf("1\n");continue;
        }
        memset(p,0,sizeof(p));
        int ans=0;
        for(int i=n;i>=1;i--)
        if(s[i]=='0') add(a[i]);
        else if(query(a[i]))
        {
            ans=1;break;
        }
        printf("%d\n",ans);
    }
}