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:
#!/usr/bin/env python3
import sys

YES = "Yes"  # type: str
NO = "No"  # type: str


def gcd(a, b):
    if b>a:
        a, b = b, a
    while b>0:
        a, b = b, a%b
    return a

def solve(T: int, A: "List[int]", B: "List[int]", C: "List[int]", D: "List[int]"):
    for i in range(T):
        a, b, c, d = A[i], B[i], C[i], D[i]
        if a<b or b>d:
            print('No')
            continue
        x = (c-a+1)%b
        g = gcd(b, d)
        print('No' if -(-x//g*g) - x + c+1 < b else 'Yes')


    return


# Generated by 1.1.3 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    T = int(next(tokens))  # type: int
    A = [int()] * (T)  # type: "List[int]" 
    B = [int()] * (T)  # type: "List[int]" 
    C = [int()] * (T)  # type: "List[int]" 
    D = [int()] * (T)  # type: "List[int]" 
    for i in range(T):
        A[i] = int(next(tokens))
        B[i] = int(next(tokens))
        C[i] = int(next(tokens))
        D[i] = int(next(tokens))
    solve(T, A, B, C, D)

if __name__ == '__main__':
    main()