Problem: 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.

Examples

Input

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


Output

No
Yes
No
Yes
Yes
No
No
Yes
No
Yes
Yes
No
No
Yes


Input

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


Output

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 JAVA):

Solution:
import java.util.*;

public class Main

{    //f[i][j]=f[i][k-1]*f[k+1][j]+f[k][k];
	static long  gcd(long  x,long  y){
		if(y==0)return x;
		return gcd(y,x%y);
	}
    public static void main(String[] args)
	{
    	 Scanner in=new  Scanner(System.in);
    long a,b,c,d;
    int t=in.nextInt() ;
    while(t-->0) {
    a=in.nextLong();
    b=in.nextLong();
    c=in.nextLong();
    d=in.nextLong();
   	if(a<b||d<b)
   	{System.out.printf("No\n");
   	  continue;
   	}
   	if(c+1>=b)
   	{
   		System.out.printf("Yes\n");
   		continue;
   	}
   	long p=gcd(b,d);
   	if((a-c-1)/p-(a-b)/p>0)
   		System.out.printf("No\n");
   	else
   		System.out.printf("Yes\n");
   	//System.out.print(p);
   }

        
     }
 
}