Problem:
Score : 600 points

Problem Statement

Ringo Mart, a convenience store, sells apple juice.

On the opening day of Ringo Mart, there were A cans of juice in stock in the morning.
Snuke buys B cans of juice here every day in the daytime.
Then, the manager checks the number of cans of juice remaining in stock every night.
If there are C or less cans, D new cans will be added to the stock by the next morning.

Determine if Snuke can buy juice indefinitely, that is, there is always B or more cans of juice in stock when he attempts to buy them.
Nobody besides Snuke buy juice at this store.

Note that each test case in this problem consists of T queries.

Constraints

1 \leq T \leq 300

1 \leq A, B, C, D \leq 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

T
A_1 B_1 C_1 D_1
A_2 B_2 C_2 D_2
:
A_T B_T C_T D_T

In the i-th query, A = A_i, B = B_i, C = C_i, D = D_i.

Output

Print T lines. The i-th line should contain Yes if Snuke can buy apple juice indefinitely in the i-th query; No otherwise.

Sample Input 1

14
9 7 5 9
9 7 6 9
14 10 7 12
14 10 8 12
14 10 9 12
14 10 7 11
14 10 8 11
14 10 9 11
9 10 5 10
10 10 5 10
11 10 5 10
16 10 5 10
1000000000000000000 17 14 999999999999999985
1000000000000000000 17 15 999999999999999985

Sample Output 1

No
Yes
No
Yes
Yes
No
No
Yes
No
Yes
Yes
No
No
Yes

In the first query, the number of cans of juice in stock changes as follows: (D represents daytime and N represents night.)

9
→D 2
→N 11
→D 4
→N 13
→D 6
→N 6
→D x

In the second query, the number of cans of juice in stock changes as follows:

9
→D 2
→N 11
→D 4
→N 13
→D 6
→N 15
→D 8
→N 8
→D 1
→N 10
→D 3
→N 12
→D 5
→N 14
→D 7
→N 7
→D 0
→N 9
→D 2
→N 11
→D …

and so on, thus Snuke can buy juice indefinitely.

Sample Input 2

24
1 2 3 4
1 2 4 3
1 3 2 4
1 3 4 2
1 4 2 3
1 4 3 2
2 1 3 4
2 1 4 3
2 3 1 4
2 3 4 1
2 4 1 3
2 4 3 1
3 1 2 4
3 1 4 2
3 2 1 4
3 2 4 1
3 4 1 2
3 4 2 1
4 1 2 3
4 1 3 2
4 2 1 3
4 2 3 1
4 3 1 2
4 3 2 1

Sample Output 2

No
No
No
No
No
No
Yes
Yes
No
No
No
No
Yes
Yes
Yes
No
No
No
Yes
Yes
Yes
No
No
No

Solution in Rust (1.15.1) (judge verdict: Time Limit Exceeded)


#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{VecDeque, HashSet, BTreeSet, BinaryHeap, HashMap};
#[allow(unused_imports)]
use std::io::{stdin, stdout, Write};


fn main() {
    let t: usize = read();
    let query = read_matrix::<i64>();

    let mut history: HashMap<(i64, i64, i64), i64> = HashMap::new();
    for i in 0..t {
        let (mut a, b, c, d) = (query[i][0], query[i][1], query[i][2], query[i][3]);
        if let Some(&i) = history.get(&(a, b, d)) {
            if i <= c {
                println!("Yes");
                continue;
            }
        }
        let a0 = a;
        loop {
            if a < b {
                println!("No");
                break;
            } else if c >= b && d >= c {
                println!("Yes");
                break;
            } else if a > c && (a - c) % b == 0 {
                if d % b == 0 {
                    println!("Yes");
                    break;
                }
                a = c + d;
            } else {
                let n = max(0, (a - c) / b);
                if a > a0 && (a - a0) % b == 0 {
                    println!("Yes");
                    break;
                } else if a - b * (n + 1) < 0 {
                    println!("No");
                    break;
                } else {
                    a = a - b * (n + 1) + d;
                    if a < c {
                        if b > d {
                            println!("No");
                            break;
                        } else if b < d {
                            if a0 > a && (a0 - a) % (d - b) == 0 {
                                println!("Yes");
                                break;
                            }
                            a = a + ((c - a) / (d - b)) * (d - b);
                        } else {
                            println!("Yes");
                            break;
                        }
                    }
                }
            }
            if a == a0 {
                println!("Yes");
                break;
            }
        }
        history.insert((a, b, d), c);
    }
}

#[allow(dead_code)]
fn read<T>() -> T where T: std::str::FromStr, T::Err: std::fmt::Debug {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    return buf
        .trim()
        .parse()
        .unwrap();
}

#[allow(dead_code)]
fn read_vector<T>() -> Vec<T> where T: std::str::FromStr, T::Err: std::fmt::Debug {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();
    return buf
        .split_whitespace()
        .map(|s| s.parse().unwrap())
        .collect();
}

#[allow(dead_code)]
fn read_matrix<T>() -> Vec<Vec<T>> where T: std::str::FromStr, T::Err: std::fmt::Debug {
    use std::io::prelude::*;
    let stdin = stdin();
    return stdin
        .lock()
        .lines()
        .map(|line| line
            .unwrap()
            .split_whitespace()
            .map(|s| s.parse().unwrap())
            .collect())
        .collect();
}