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>
using namespace std;
template<class T> ostream& operator<<(ostream &os, vector<T> V) { os << "[ "; for(auto v : V) os << v << " "; return os << "]"; }
template<class T> ostream& operator<<(ostream &os, set<T> S){ os << "{ "; for(auto s:S) os<<s<<" ";return os<<"}"; }
template<class L, class R> ostream& operator<<(ostream &os, pair<L,R> P) { return os << "(" << P.first << "," << P.second << ")";}
template<class L, class R> ostream& operator<<(ostream &os, map<L,R> M) { os << "{ "; for(auto m:M) os<<"("<<m.first<<":"<<m.second<<") "; return os<<"}"; }
#ifdef LMAO
    #define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
    template <typename Arg1>
    void __f(const char* name, Arg1&& arg1){cerr << name << " : " << arg1 << endl;}
    template <typename Arg1, typename... Args>
    void __f(const char* names, Arg1&& arg1, Args&&... args){ const char* comma = strchr(names + 1, ','); cerr.write(names, comma - names) << " : " << arg1<<" | "; __f(comma+1, args...); }
#else
    #define trace(...) 42
#endif
#define ff first
#define ss second
#define pb push_back
#define eb emplace_back
#define endl "\n"
#define tcase int tt; cin>>tt; while(tt--)
#define tcase2 int tt; cin>>tt; for(int qq=1;qq<=tt;qq++)
#define all(x) x.begin(), x.end()
typedef long long int ll;
typedef  unsigned long long int ull;
typedef long double ld;
const int MAX = 1000010;
const int MOD2 = 998244353;
const int MOD = 1000000007; 
const int INF = 1e8;    
const ll LLINF = 0x3f3f3f3f3f3f3f3f;
const ld EPS = 1e-7;
#define forn(i, n) for(int i = 0; i < (int)n; i++)
#define forne(i, a, b) for(int i = a; i <= b; i++)   
/*****************start************/
void solve()
{
    int n;
    cin>>n;
    vector<ll> a(n);
    map<int,int> fa,fb;
    for(auto &i:a)
        cin>>i;
    string s;
    cin>>s;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<64;j++)
        {
            int lo=((a[i]>>j)&1);
            if(s[i]=='0')
                fa[j]+=lo;
            else
                fb[j]+=lo;
        }
    }
    for(int i=0;i<64;i++)
    {
        if(fa[i]<fb[i])
        {
            cout<<1<<endl;
            return;
        }
    }
    cout<<0<<endl;
}
int main()
{
    ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);cout<<fixed<<setprecision(25);cerr<<fixed<<setprecision(10);
    auto start = std::chrono::high_resolution_clock::now();
    tcase{
        solve();
    }
    auto stop = std::chrono::high_resolution_clock::now(); 
    auto duration = std::chrono::duration_cast<std::chrono::nanoseconds>(stop - start); 
    // cerr << "Time taken : " << ((long double)duration.count())/((long double) 1e9) <<"s "<< endl;     
    return 0;
}