Problem: 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.

Example

Input

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


Output

1
0
0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define ALL(v) (v).begin(), (v).end()
using ll = long long;
using P = pair<int, int>;
constexpr int INF = 1e9;
constexpr long long LINF = 1e18;
constexpr long long MOD = 1e9 + 7;

void solve() {
    int n;
    cin >> n;
    ll a[n];
    rep(i, n) {
        cin >> a[i];
    }
    string s;
    cin >> s;
    reverse(a, a + n);
    reverse(ALL(s));
    vector<ll> bs;
    rep(i, n) {
        for (auto b : bs) {
            a[i] = min(a[i], a[i] ^ b);
        }
        if (s[i] == '0') {
            if (a[i] != 0) bs.emplace_back(a[i]);
        } else {
            if (a[i]) {
                cout << 1 << endl;
                return;
            }
        }
    }
    cout << 0 << endl;
}

signed main() {
    int tt;
    cin >> tt;
    while (tt--) {
        solve();
    }
    return 0;
}