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: Accepted)

use proconio::{input, marker::Chars, source::line::LineSource};
use std::io::stdin;

fn main() {
    let stdin = stdin();
    let mut src = LineSource::new(stdin.lock());
    input! {
        from &mut src,
        t: usize,
    }
    for _ in 0..t {
        input! {
        from &mut src,
            n: usize,
            aa: [u64; n],
            s: Chars
        }
        solve(aa, s);
    }
}

fn solve(aa: Vec<u64>, s: Vec<char>) {
    let mut dp: Vec<u64> = vec![];
    for (mut a, c) in aa.into_iter().zip(s.into_iter()).rev() {
        for &v in dp.iter() {
            let t = 1 << v.trailing_zeros();
            if t & a != 0 {
                a ^= v;
            }
        }
        if c == '0' {
            if a != 0 {
                dp.push(a);
            }
        } else {
            if a != 0 {
                println!("1");
                return;
            }
        }
    }
    println!("0");
}