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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Numerics;
using System.Text;
using System.Threading.Tasks;
using static System.Console;
using static System.Math;
class Z { static void Main() => new K(); }
class K
{
	int F => int.Parse(Str);
	long FL => long.Parse(Str);
	int[] G => Strs.Select(int.Parse).ToArray();
	long[] GL => Strs.Select(long.Parse).ToArray();
	string Str => ReadLine();
	string[] Strs => Str.Split(new char[] { ' ', '\t' }, StringSplitOptions.RemoveEmptyEntries);
	const int MOD = 1000000007;
	public K()
	{
		SetOut(new StreamWriter(OpenStandardOutput()) { AutoFlush = false });
		Solve();
		Out.Flush();
	}
	void Solve()
	{
		var T = F;
		for (int i = 0; i < T; i++)
		{
			var I = GL;
			WriteLine(Check(I[0], I[1], I[2], I[3]) ? "Yes" : "No");
		}
	}
	bool Check(long A, long B, long C, long D)
	{
		if (A < B || D < B) return false;
		if (C >= B) return true;
		var g = GCD(B, D);
		return A % g + B - g <= C;
	}
	public static long GCD(long a, long b)
	{
		var n = (ulong)Math.Abs(a); var m = (ulong)Math.Abs(b);
		if (n == 0) return (long)m; if (m == 0) return (long)n;
		int zm = 0, zn = 0;
		while ((n & 1) == 0) { n >>= 1; zn++; }
		while ((m & 1) == 0) { m >>= 1; zm++; }
		while (m != n)
		{
			if (m > n) { m -= n; while ((m & 1) == 0) m >>= 1; }
			else { n -= m; while ((n & 1) == 0) n >>= 1; }
		}
		return (long)n << Math.Min(zm, zn);
	}
}