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:
T = int(input())

ABCD = []
for i in range(T):
    a,b,c,d = map(int,input().split())
    ABCD.append((a,b,c,d))

for abcd in ABCD:
    a,b,c,d = abcd
    # 初日にダメになる
    if a < b:
        print("No")
    # 在庫補充が追いつかない
    elif d < b:
        print("No")
    # 補充タイミングが十分に早い
    elif c >= b:
        print("Yes")
    else:
        #print(a,b,c,d,a%b,d%b)
        e = a%b
        f = d%b
        list = [a]
        if f==0:
            if c<e<b:
                print("No")
            else:
                print("Yes") 
        else:
            cnt = 0
            while cnt < 100:
                g = ((c-e)//f+1)*f+e
                if g < b:
                    print("No")
                    cnt += 10000
                else:
                    if g in list:
                        print("Yes")
                        cnt += 10000
                    else:
                        list.append(g)
                        a = g
                        e = a%b
                        f = d%b
                        cnt += 1