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: Wrong Answer)

#include<bits/stdc++.h>
#include <ext/rope>

#define min3(a,b,c)   min(a,min(b,c))
#define max3(a,b,c)   max(a,max(b,c))
#define min4(a,b,c,d) min(min(a,b),min(c,d))
#define max4(a,b,c,d) max(max(a,b),max(c,d))

#define count_one(a) __builtin_popcount(a)  // Returns the number of set bits(1) in a number(a). In long long use __builtin_popcountll(a)
#define parity(i)    __builtin_parity(i)  //even parity 0 and odd parity 1
#define blz(a)       __builtin_clz(a) //Returns the number of leading zeroes in a number(a)
#define btz(a)       __builtin_ctz(a) //Returns the number of trailing zeroes in a number(a)
#define gcd(a,b)     __gcd(a,b)
#define lcm(a,b)     (a*(b/gcd(a,b)))

#define pb   push_back
#define PI   M_PI
#define endl "\n"
#define sc   scanf
#define pf   printf

#define ll  long long
#define ull unsigned long long

#define W(t)            while(t--)
#define rep1(i,n)       for(int i = 0; i < n; i++)
#define rep2(a,b)       for(int i = a; i <= b; i++)
#define rep3(a,b,c)     for(int i = a; i <= b; i+=c)
#define irep(a,b,c)     for(int i = b; i >=a; i-=c)
#define repit(it,type)  for(it = type.begin(); it != type.end(); it++)
#define IOS ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
using namespace std;
using namespace __gnu_cxx;

int main()
{
    IOS
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    //cout<<(2^3);

    ll t,n,i;
    string s;

    cin>>t;

    W(t)
    {
         cin>>n;
         ll a[n],b[n]={};
         ll x = 0;
         rep1(i,n)
         {
              cin>>a[i];
         }
         cin>>s;

         rep1(i,s.size())
         {
              if(s[i]=='1')
              {
                   if((x!=a[i] && x+1!=a[i])||!x)
                    x = x^a[i];
              }
              else
              {
                   if(x && x!=a[i+1])
                    x = x^a[i];
              }
         }
         if(x)
          cout<<1<<endl;
         else
          cout<<0<<endl;

    }

    return 0;
}