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: Wrong Answer)

#[allow(unused_macros)]
macro_rules! parse_line {
    ( $t:ty ) => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            iter.next().unwrap().parse::<$t>().unwrap()
        }
    );

    ( $( $t:ty), +) => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            ( $(iter.next().unwrap().parse::<$t>().unwrap()),* )
        }
    );
}

#[allow(unused_macros)]
macro_rules! read_line {
    () => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            line.pop();
            line
        }
    )
}

#[allow(unused_macros)]
macro_rules! parse_vec {
    ( $t:ty ) => (
        {
            let mut line = String::new();
            ::std::io::stdin().read_line(&mut line).unwrap();
            let iter = line.split_whitespace();
            iter.map(|v| v.parse::<$t>().unwrap()).collect::<Vec<_>>()
        }
    )
}

fn solve(a: u64, b: u64, c: u64, d: u64) -> bool {
    use std::collections::HashMap;

    if a < b {
        return false
    }

    if d < b {
        return false
    }

    if b-1 <= c {
        return true
    }

    let mut m = HashMap::new();
    m.insert(a, true);

    let mut na = a;

    let dmodb = if d % b == 0 { d } else {d % b};
    let mut r = na % b;

    while true {
        if r > c {
            return false
        }

        let na = c + (((c - r) / dmodb ) + 1) * dmodb;
        if na > c && na < b {
            return false
        }
        if m.contains_key(&na) {
            return true
        }

        m.insert(na, true);
        r = na % b;
    }

    unreachable!()
}

fn main() {
    let n = parse_line!(usize);
    let mut ss = Vec::new();

    for _ in 0..n {
        ss.push(parse_vec!(u64));
    }

    for v in ss {
        println!("{}", solve(v[0], v[1], v[2], v[3]));
    }
}

#[test]
fn solve_test() {
    assert_eq!(solve(7,7,5,7), true);
    assert_eq!(solve(13,7,5,7), false);
    assert_eq!(solve(7,2,1,3), true);
    assert_eq!(solve(1000000000000000000, 17, 14, 999999999999999985), false);
    assert_eq!(solve(1000000000000000000, 17, 15, 999999999999999985), true);
}