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 PYTHON3):

Solution:
def gcd(a, b):
	while b:
		a, b = b, a % b
	return a

T=int(input())
Juice=[None]*T
for i in range(T):
    Juice[i]=list(map(int,input().split()))

#A最初の在庫,Bすぬけ君が買う本数,C以下の場合D追加


#A→A%B
def juicecheck(i):
    A=Juice[i][0]
    B=Juice[i][1]
    C=Juice[i][2]
    D=Juice[i][3]

    if A<B:
        print("No")
        return
    elif B>D:
        print("No")
        return
    elif B-1<=C:
        print("Yes")
        return
    else:
        x=A%B
        y=D%B
        

        z=gcd(B,D)

        if z==1 or B-(C+1)>=z or  (C+1)%z<=x%z<=(B-1)%z or ( (C+1)%z>B%z and ((C+1)%z<=x%z or x%z<B%z) ):
            print("No")
            return

        print("Yes")
        return
                
        #(x+n*y)%B==[C+1,B)となるnがあれば"No"

for i in range(T):
    juicecheck(i)