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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.PrintStream;
import java.util.Scanner;
public class Main {
    private static final PrintStream so = System.out;
    private static final Scanner     sc = new Scanner(System.in);

    public static void main(String[] args) {
        int t = sc.nextInt();
        

        outer: for (int i = 0; i < t; i++) {
            long a = sc.nextLong();
            long b = sc.nextLong();
            long c = sc.nextLong();
            long d = sc.nextLong();
            
            if (a < b) {
                so.println("No");
                continue;
            }
            if (b > d) {
                so.println("No");
                continue;
            }
            if (b == d) {
                so.println((a % b <= c) ? "Yes" : "No");
                continue;
            }
            if (a % b > c) {
                so.println("No");
                continue;
            }
            
            long supAB = a % b;
            final long diffDB = d - b;
            
            for (int j = 0; j < 500000; j++) {
                long next =  f(c, supAB, diffDB);
//                so.println("c: " + c + ", i: " + i + ", sup: " + supAB + ", next: " + next);
                if (next < b) {
                    so.println("No");
                    continue outer;
                }
                if (next + d < b) {
                    so.println("No");
                    continue outer;
                }
                supAB = (next + d) % b;
//                so.println("c: " + c + ", i: " + i + ", sup: " + supAB + ", next: " + next);
//                so.println();
            }
            so.println("Yes");
        }
    }
    
    static long f(long x, long sup, long diff) {
        long temp = ((x - sup - 1) / diff + 1) * diff + sup;
        if (x == temp) temp += diff;
        return temp;
    }
}
/*
1
9 7 5 9


*/