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.awt.Point;
import java.util.ArrayDeque;
import java.util.Scanner;
import java.util.*;
import java.util.Arrays;
import java.util.ArrayList;
import java.util.Set;
import java.util.HashSet;
import java.util.Collection;


 
public class Main{
	
        static final Scanner sc=new Scanner(System.in);
        
        
        
        
        
        
        public static void main(String[]args){
                int T=sc.nextInt();
                for(int i=0;i<T;i++){
                        long A=sc.nextLong();
                        long B=sc.nextLong();
                        long C=sc.nextLong();
                        long D=sc.nextLong();
                        long E=D-B;
                        long hozon=0;
                        ArrayList<Long>kazoe=new ArrayList<>();
                        if(A<B){
                                System.out.println("No");
                        }else if((E>0)&&(A+E*5000<C)){System.out.println("Yes");}
                        else{
                                
                                        
                                
                                
                                for(;;){if(kazoe.size()>1000){
                                        long F=C-A;
                                        F=F/E;
                                        A=A+F*E;
                                        
                                }
                                if(kazoe.size()>6000){
                                        System.out.println("Yes");
                                        break;
                                }
                                        
                                        long X=A/B-1;
                                        if(X>0){
                                                A=A-B*X;

                                        }
                                        if(A<B){
                                                System.out.println("No");
                                                break;
                                        }else{
                                        if(A-B<=C){
                                                A=A-B+D;
                                                if(kazoe.contains(A)){
                                                        System.out.println("Yes");
                                                        break;
                                                }
                                                kazoe.add(A);
                                                hozon=A;
                                               
                                                
                                        }
                                        else{
                                                A=A-B;
                                                if(kazoe.contains(A)){
                                                        System.out.println("Yes");
                                                        break;
                                                }
                                                kazoe.add(A);

                                        }
                                       
                                }

                                }
                        }
                }
                
                


                
        }
        
        
        

		
	
}