Problem:
Score : 600 points

Problem Statement

Let us consider the following operations on a string consisting of A and B:

Select a character in a string. If it is A, replace it with BB. If it is B, replace with AA.

Select a substring that is equal to either AAA or BBB, and delete it from the string.

For example, if the first operation is performed on ABA and the first character is selected, the string becomes BBBA.
If the second operation is performed on BBBAAAA and the fourth through sixth characters are selected, the string becomes BBBA.

These operations can be performed any number of times, in any order.

You are given two string S and T, and q queries a_i, b_i, c_i, d_i.
For each query, determine whether S_{a_i} S_{{a_i}+1} ... S_{b_i}, a substring of S, can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, a substring of T.

Constraints

1 \leq |S|, |T| \leq 10^5

S and T consist of letters A and B.

1 \leq q \leq 10^5

1 \leq a_i \leq b_i \leq |S|

1 \leq c_i \leq d_i \leq |T|

Input

Input is given from Standard Input in the following format:

S
T
q
a_1 b_1 c_1 d_1
...
a_q b_q c_q d_q

Output

Print q lines. The i-th line should contain the response to the i-th query. If S_{a_i} S_{{a_i}+1} ... S_{b_i} can be made into T_{c_i} T_{{c_i}+1} ... T_{d_i}, print YES. Otherwise, print NO.

Sample Input 1

BBBAAAABA
BBBBA
4
7 9 2 5
7 9 1 4
1 7 2 5
1 7 2 4

Sample Output 1

YES
NO
YES
NO

The first query asks whether the string ABA can be made into BBBA.
As explained in the problem statement, it can be done by the first operation.

The second query asks whether ABA can be made into BBBB, and the fourth query asks whether BBBAAAA can be made into BBB.
Neither is possible.

The third query asks whether the string BBBAAAA can be made into BBBA.
As explained in the problem statement, it can be done by the second operation.

Sample Input 2

AAAAABBBBAAABBBBAAAA
BBBBAAABBBBBBAAAAABB
10
2 15 2 13
2 13 6 16
1 13 2 20
4 20 3 20
1 18 9 19
2 14 1 11
3 20 3 15
6 16 1 17
4 18 8 20
7 20 3 14

Sample Output 2

YES
YES
YES
YES
YES
YES
NO
NO
NO
NO

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

#![allow(non_snake_case)]
#![allow(unused_variables)]

fn main() {
    let S: Vec<char> = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.trim().chars().collect()
    };
    let T: Vec<char> = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.trim().chars().collect()
    };
    let q: usize = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.trim().parse().unwrap()
    };
    let (a, b, c, d): (Vec<usize>, Vec<usize>, Vec<usize>, Vec<usize>) = {
        let (mut a, mut b, mut c, mut d) = (vec![], vec![], vec![], vec![]);
        for _ in 0..q {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            a.push(iter.next().unwrap().parse().unwrap());
            b.push(iter.next().unwrap().parse().unwrap());
            c.push(iter.next().unwrap().parse().unwrap());
            d.push(iter.next().unwrap().parse().unwrap());
        }
        (a, b, c, d)
    };

    // 並び替えは任意に可能
    // A, Bの個数は+1/+1または+3/+0または+0/+3できることを利用する
    let mut acc_S = vec![0; S.len() + 1];
    let mut acc_T = vec![0; T.len() + 1];
    for i in 0..(S.len()) {
        acc_S[i + 1] = if S[i] == 'A' {
            acc_S[i] + 1
        } else {
            acc_S[i]
        };
    }
    for i in 0..(T.len()) {
        acc_T[i + 1] += if T[i] == 'A' {
            acc_T[i] + 1
        } else {
            acc_T[i]
        };
    }
    let ans = (0..q).map(|i| {
            let a_s = (acc_S[b[i]] - acc_S[a[i] - 1]) as i64;
            let b_s = ((b[i] - a[i] + 1) as i64) - a_s;
            let a_t = (acc_T[d[i]] - acc_T[c[i] - 1]) as i64;
            let b_t = ((d[i] - c[i] + 1) as i64) - a_t;
            println!("{} {} {} {}", a_s, b_s, a_t, b_t);
            if b_t % 3 == ((b_s + (a_t - a_s))) % 3 {
                "YES"
            } else {
                "NO"
            }
        })
        .collect::<Vec<_>>()
        .join("\n");

    println!("{}", ans);
}