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_imports)]
use std::cmp::{min,max};
#[allow(unused_imports)]
use std::collections::BTreeMap;
#[allow(unused_imports)]
use std::ops::*;
#[allow(unused_imports)]
use std::collections::BinaryHeap;

fn gcd(x: i64, y: i64) -> i64 {
    if y == 0 {
        x
    }
    else {
        gcd(y, x%y)
    }
}

fn sim(a: i64, b: i64, c: i64, d: i64) -> bool {
    unimplemented!();
}

fn main() {
    let n = read_i64() as usize;
    for _ in 0..n {
        let (a,b,c,d) = {
            let xs = read_vec_i64();
            (xs[0], xs[1], xs[2], xs[3])
        };
        let res = if a < b {
            false
        }
        else if b > d {
            false
        }
        else if b <= c && b <= d {
            true
        }
        else if c < b && b == d {
            0 <= a % b && a % b <= c
        }
        else if c + 1 == b && b < d {
            true
        }
        else if c < b && b < d {
            //println!("c={}, a-b={}, d-b={}, b={}, a%b={}", c, a-b, d-b, b, a%b);
            //println!("??={}", (c-((d%b)-b)+10*b)%b);
            //let t = ((b-a%b/(d-b)*(d-b))%b;
            //println!("t={}", t);
            if a%b < c && (b-1-a%b)/(d-b) != (c-a%b)/(d-b) {
                false
            }
            else {
                0 <= a % b && a % b <= c
            }
        }
        else {
            unreachable!();
        };
        println!("{}", if res { "Yes" } else { "No" });
        //println!();
    }
}

#[allow(dead_code)]
fn read_line() -> String {
    let mut ret = String::new();
    std::io::stdin().read_line(&mut ret).ok();
    ret.pop();
    return ret;
}

#[allow(dead_code)]
fn read_i64() -> i64 {
    let ss = read_line();
    return ss.parse::<i64>().unwrap();
}

#[allow(dead_code)]
fn read_vec_i64() -> Vec<i64> {
    let mut res = vec![];
    let ss = read_line();
    for ts in ss.split_whitespace() {
        let x = ts.parse::<i64>().unwrap();
        res.push(x);
    }
    return res;
}

use std::fmt::Display;
#[allow(dead_code)]
fn write_vec<T: Display>(xs: &Vec<T>) {
    if xs.len() == 0 {
        println!("");
        return;
    }
    print!("{}", xs[0]);
    for i in 1..xs.len() {
        print!(" {}", xs[i]);
    }
    println!("");
}