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::stdin;

fn read_line() -> Vec<String> {
    let mut line = String::new();
    stdin().read_line(&mut line).expect("Failed to read line");
    return line.trim().split_whitespace().map(|s| s.to_string()).collect();
}

struct Reader {
    v: Vec<String>,
    n: usize,
}

impl Reader {
    fn new() -> Reader {
        let v = read_line();
        return Reader { v: v, n: 0 };
    }
    fn next<T>(&mut self) -> T where T: std::str::FromStr, T::Err: std::fmt::Debug {
        if self.n == self.v.len() {
            self.v = read_line();
            self.n = 0;
        }
        let x = self.v[self.n].parse().expect(&self.v[self.n]);
        self.n += 1;
        return x;
    }
    fn nexts<T>(&mut self, n: usize) -> Vec<T>
    where T: std::str::FromStr, T::Err: std::fmt::Debug {
        let mut v = Vec::with_capacity(n);
        for _ in 0..n {
            v.push(self.next());
        }
        return v;
    }
}

fn gcd(x: u64, y: u64) -> u64 {
    if y==0 {
        return x;
    } else {
        return gcd(y, x%y);
    }
}

fn main() {
    let mut r = Reader::new();
    let t: usize = r.next();
    for _ in 0..t {
        let a: u64 = r.next();
        let b: u64 = r.next();
        let c: u64 = r.next();
        let d: u64 = r.next();
        if a < b {
            println!("No");
            continue;
        }
        if d < b {
            println!("No");
            continue;
        }
        if c >= b {
            println!("Yes");
            continue;
        }

        let a = a % b;
        let d = d % b;
        
        let g = gcd(d, b);
        let k = (b - a - 1) / g;
        if a + k * g <= c {
            println!("Yes");
        } else {
            println!("No");
        }
    }
}