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 (GCC 5.4.1) (judge verdict: Accepted)

int main(void) {
int  t, i;
	long long reft,r,x,y,s,k;
	scanf("%d", &t);
	long long  a[300], b[300], c[300], d[300];
	for (i = 0; i <= t-1; i++) {
		scanf("%lld", &a[i]);
		scanf("%lld", &b[i]);
		scanf("%lld", &c[i]);
		scanf("%lld", &d[i]);
 
	}
 
	for (i = 0; i <= t-1; i++) {
		if (a[i] < b[i]) printf("No\n");
		else if (b[i] > d[i]) printf("No\n");
		else if (b[i]-1 <= c[i]) printf("Yes\n");
		else {	
				reft = a[i]%b[i];
			
			if (reft > c[i]) printf("No\n");
			else {
								x = d[i];
				y = b[i];
				
				r = x % y;
				while (r != 0) {
					x = y;
					y = r;
					r = x % y;
				}
				
				c[i]=c[i]-reft;
                                b[i]=b[i]-reft;
                                k=c[i]%y;
                                s=(c[i]-k)/y;
                                
 
 
				if ((s+1)*y<b[i]) printf("No\n");
				else printf("Yes\n");
 
			}
 
		}
 
	}
	return 0;
}