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 Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::input;
use proconio::marker::{Chars};

// x ** n % mods を求める
pub fn mod_pow(x: i64, n: i64, mods: i64) -> i64 {
    if n == 0 {
        return 1;
    }
    let mut res = mod_pow(x * x % mods, n / 2, mods);
    if n % 2 == 1 {
        res = res * x % mods;
    }
    return res;
}

// (1/n) % mods を求める
pub fn mod_inv(n: i64, mods: i64) -> i64 {
    return mod_pow(n, mods-2, mods);
}

// nCk % mods を求める
pub fn mod_comb(n: i64, k: i64, mods: i64) -> i64 {
    let sum1 = (n-k+1..n+1).fold(1, |ac, x| ac * x % mods);
    let sum2 = (1..k+1).fold(1, |ac, x| ac * x % mods);
    return sum1 * mod_inv(sum2, mods) % mods;
}

// nCk の組み合わせを列挙する
pub fn combination(n: usize, k: usize) -> Vec<Vec<usize>> {
    let mut ans = Vec::new();
    if k == 1 {
        for i in 1..n+1 {
            let mut tmp = Vec::new();
            tmp.push(i);
            ans.push(tmp);
        }
    }
    else {
        for iter in combination(n-1, k-1) {
            let max = iter.iter().max().unwrap();
            for j in *max+1..n+1 {
                let mut tmp = iter.clone();
                tmp.push(j);
                ans.push(tmp);
            }
        }
    }
    ans
}


fn main() {
    input!{
        t: usize,
    }
    for _ in 0..t {
        input!{
            n: usize,
            a: [usize; n],
            s: Chars,
        }
        if s[n-1] == '1' {
            println!("1");
        }
        else {
            let mut dp0 = vec![0; 60];
            let mut dp1 = vec![0; 60];
            for i in 0..n {
                let tmp = format!("{:b}", a[i]);
                //            println!("{:?}", tmp);
                if s[i] == '0' {
                    for j in 0..tmp.len() {
                        dp0[j] += tmp.chars().rev().nth(j).unwrap() as i64 - 48;
                    }
                }
                else {
                    for j in 0..tmp.len() {
                        dp1[j] += tmp.chars().rev().nth(j).unwrap() as i64 - 48;
                    }
                }
            }
            // println!("{:?}", dp0);
            // println!("{:?}", dp1);
            let mut flag = 0;
            for i in 0..60 {
                if dp1[i] > dp0[i] {
                    flag = 1;
                    break;
                }
            }
            println!("{}", flag);
        }
    }
}