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 {
	static long gcd(long a,long b) {
		if(b == 0)return a;
		return gcd(b,a % b);
	}
	static boolean check(long a,long b,long c,long d) {
		if(a < b || d < b)return false;
		if(c + 1 >= b)return true;
		///如果某天之后  剩下的多于c(不需要增加）  但是却小于b 那么次日肯定不够 
		///实际上就是c < a - bx + dy < b有解的话就是No 
		///移项的a - b < bx - dy < a - c
		///根据欧几里德扩展定理 就是说(a-b,a-c)之间是否有gcd(b,d)的倍数 即是否有解
		long g = gcd(b,d);
		return (a - c - 1) / g - (a - b) / g <= 0;
	}
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		for(int i = 0;i < t;i ++) {
			long a = in.nextLong();
			long b = in.nextLong();
			long c = in.nextLong();
			long d = in.nextLong();
			if(check(a,b,c,d))System.out.println("Yes");
			else System.out.println("No");
		}
	}

}