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: Time Limit Exceeded)

import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

public class Main {
	static int i;
	static long []A;
	static long []B;
	static long []C;
	static long []D;
	static StringBuilder sb;
	// 在庫
	static long a;
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int T = scan.nextInt();
		A = new long[T];
		B = new long[T];
		C = new long[T];
		D = new long[T];

		for(i = 0; i < T; i++) {
			A[i] = scan.nextLong();
			B[i] = scan.nextLong();
			C[i] = scan.nextLong();
			D[i] = scan.nextLong();
		}
		scan.close();
		sb = new StringBuilder();
		for(i = 0; i < T; i++) {
			a = A[i] - B[i];
			if(a < 0 || D[i] < B[i]) {
				sb.append("No").append("\n");
			}else if(C[i] >= B[i]){
				sb.append("Yes").append("\n");
			}else {
				// C > B, A > B, D > B, a > 0
				Set<Long> map = new HashSet<Long>();
				map.add(a);
				int cnt = 0;
				while(true) {
					if(isA()) {
						break;
					}
					if(a <= C[i]) {
						if(map.contains(a)) {
							cnt ++;
							if(cnt >= 2) {
								sb.append("Yes").append("\n");
								break;
							}
						}else {
							map.add(a);
						}
					}
					add();
					a -= B[i];
					if(isA()) {
						break;
					}
					while(a > C[i]) {
						long l = a - C[i];
						long k = l / B[i];
						if(k == 0) {
							a -= B[i];
						}else {
							a -= k * B[i];
						}
					}
				}
			}

		}
		System.out.println(sb.toString());

	}
	// 在庫を補充
	static void add() {
		if(a <= C[i]) {
			a += D[i];
		}
	}
	static boolean isA() {
		if(a < 0) {
			sb.append("No").append("\n");
			return true;
		}else {
			return false;
		}
	}
}