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: Accepted)

use std::io::*;
use std::str::FromStr;

#[allow(dead_code)]
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

#[allow(dead_code)]
fn naive(a: i64, b: i64, c: i64, d: i64) -> bool {
    let mut cnt = a;
    let mut visited = vec![false; (a + c + d + 1) as usize];
    while !visited[cnt as usize] {
        println!("{}", cnt);
        visited[cnt as usize] = true;
        if cnt < b {
            return false;
        }
        cnt -= b;
        if cnt <= c {
            cnt += d;
        }
    }
    return true;
}

#[allow(dead_code)]
fn gcd(a: i64, b: i64) -> i64 {
    if b == 0 {
        return a;
    }
    gcd(b, a % b)
}

fn solve(a: i64, b: i64, c: i64, d: i64) -> bool {
    if a < b || d < b {
        return false;
    }
    if a <= c || b <= c {
        return true;
    }
    let offset = a - (a - c) / b * b - c;
    let mut diff = d - b;
    if diff == 0 {
        return c + offset >= b;
    }
    assert!(offset >= 0);
    let g = gcd(d, diff);
    let mut min_value = c + offset % g;
    if min_value == c {
        min_value += g;
    }
    // println!("{} {} {} {} {}", diff, offset, g, offset % g, min_value);
    min_value >= b
}

fn main() {
    let t = read::<usize>();
    for _ in 0..t {
        let a = read::<i64>();
        let b = read::<i64>();
        let c = read::<i64>();
        let d = read::<i64>();
        // assert!(naive(a, b, c, d) == solve(a, b, c, d));
        // println!("{}", if naive(a, b, c, d) { "Yes" } else { "No" });
        println!("{}", if solve(a, b, c, d) { "Yes" } else { "No" });
    }
}