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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
#define fi(a) scanf("%d",&a);
#define fli(a) scanf("%ld",&a);
#define fll(a) scanf("%lld",&a);
#define pi(a) printf("%d\n",a);
#define ppi(i,a) printf("Case %d: %d\n",i,a);
#define ll long long
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
ll arr[205];
int main()
{
    fast
    int n,m,t,ans=0;
    string s;
    ll x;
    cin >> t;
    while(t--)
    {
        ans = 0;
        cin >> n;
        for(int i=1; i<=n; i++)
        {
            cin >> arr[i];
        }
        cin >> s;
        if(n==1)
        {
            int a;
            if(s[arr[1]-1]=='1')
                a=1;
            else
                a = 0;

            cout << 1 << endl;
        }
        else
        {
            int a;
            if(s[arr[1]-1]=='1')
                a=1;
            else
                a = 0;

            int b;
            if(s[arr[2]-1]=='1')
                b=1;
            else
                b = 0;

            ans = a^ans;
            ans = b^ans;
        }
        cout << ans << endl;
    }
    return 0;
}