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.*;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		// 整数の入力
		int t = sc.nextInt();
		// スペース区切りの整数の入力
		for(int i=1;i<=t;i++){

			long a = sc.nextLong();
			long b = sc.nextLong();
			long c = sc.nextLong();
			long d = sc.nextLong();
			//最初在庫A弧ある
			//B個毎日減る
			//もしC個以下なら補充
			//D個足される

			//そもそもD＜BならNo
			//A<B No
			//C＞＝BならYes
			if(a<b){
				// 初日で在庫尽きる
				System.out.println("No");
			}else if(d<b){
				// 補充が追いつかないならいつかダメ
				System.out.println("No");
			}else if(c>=b){
				// 発注点が余裕あるから大丈夫
				System.out.println("Yes");
			}else{
				// 発注点が消費量より小さい場合
			//D=Bの場合、
				// AをBで割った余りは、マイナスになる一歩手前の在庫数
				// これが発注点を超えているか否かで判定可能
				if(d==b){
					long remainQ = a%b;
				if(remainQ<=c){
					System.out.println("Yes");
				}else{
					System.out.println("No");
				}
			}else{
								//D>Bの場合、発注点近くまで近づいてから判定
			// つまりある程度在庫を削ってから判定する

				// 任意のタイミングでgoods>cかつgoods<bとなる時があるはず
				// そのタイミングが出るならNo,全くでないならYes

				long goods = a;
				while(goods>c){
					goods = goods -b;
// System.out.println("goodsRemain" + goods);
				}

				// System.out.println("zaiko" + goods);
				if(goods<0L){
					goods = goods+b;
				}
				// 	if(goods<=c){
				// 		System.out.println("Yes");
				// 	}else{
				// 		System.out.println("No");
				// 	}
				// }else{
					if(goods<=c){
						System.out.println("Yes");
					}else{
						System.out.println("No");
					}
				// }
			}

				
			}

						// System.out.println(i);
		}


	}
}