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

public class Main {
	long a, b, c, d;
	Set<Long> counts = new HashSet<>();

	Main(long a, long b, long c, long d) {
		this.a = a;
		this.b = b;
		this.c = c;
		this.d = d;
	}

	boolean simulate() {
		if (b > d || a < b) {
			// そもそも持続不可能
			return false;
		}
		long count = a % b;
		counts.add(count);
		boolean result = false;
		while (true) {
			// 夜のフェーズ
			if (count <= c) {
				count += d;
			}
			// 昼のフェーズ
			if (count >= b) {
				count = count % b;
			} else {
				result = false;
				break;
			}
			// この段階でジャッジ
			if (counts.contains(count)) {
				result = true;
				break;
			}
			counts.add(count);
			// System.out.println(count);
		}
		return result;
	}

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int t = Integer.parseInt(in.nextLine());
		long[] a = new long[t];
		long[] b = new long[t];
		long[] c = new long[t];
		long[] d = new long[t];
		for (int i = 0; i < t; ++i) {
			String[] tokens = in.nextLine().split(" ");
			a[i] = Long.parseLong(tokens[0]);
			b[i] = Long.parseLong(tokens[1]);
			c[i] = Long.parseLong(tokens[2]);
			d[i] = Long.parseLong(tokens[3]);
		}
		for (int i = 0; i < t; ++i) {
			// Main ins = new Main(a[i], b[i], c[i], d[i]);

			if (calc(a[i], b[i], c[i], d[i])) {
				System.out.println("Yes");
			} else {
				System.out.println("No");
			}
		}
		in.close();

	}

	private static boolean calc(long a, long b, long c, long d) {
		if (b <= d && a >= b) {
			if (b == d) {
				if (a % b > c && c < b) {
					return false;
				} else {
					return true;
				}
			} else {
				long start = (long) Math.floor((double) (c - a % b) / (double) (d - b));
				start *= (d - b);
				start -= (d - b);
				for (long i = a % b + start; i < b; i = i + (d - b)) {
					if (c < i && i < b) {
						return false;
					}
				}
				return true;
			}
		}
		return false;
	}

	private static long gcd(long m, long n) {
		if (m < n)
			return gcd(n, m);
		if (n == 0)
			return m;
		return gcd(n, m % n);
	}

}