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
def examA():
    N = I()
    A = LI()
    ans = 0
    for i in range(N-1):
        if A[i+1]==A[i]:
            A[i+1] = 101
            ans +=1
    print(ans)
    return

def gcd(x,y):
    if y==0:
        return x
    while(y!=0):
        x, y = y,x%y
    return x
def examB():
    T = I()
    for i in range(T):
        ans = "Yes"
        A, B, C, D = LI()
        if A<B:
            ans = "No"
        elif D<B:
            ans = "No"
        elif C>=B:
            print(ans)
            continue
        elif B-C>gcd(D,B):
            ans = "No"
        if (A % B) > C:
            ans = "No"
        print(ans)
    return

def examC():
    N = I()
    S = SI()
    d1 = defaultdict(int)
    d2 = defaultdict(int)
    loop = 2**N
    for i in range(loop):
        curR = ""; curB = ""
        curR2 = ""; curB2 = ""
        for j in range(N):
            if i&(1<<j)==(1<<j):
                curR +=S[j]
                curR2 +=S[j+N]
            else:
                curB +=S[j]
                curB2 +=S[j+N]
        d1[(curR,curB)] +=1
        d2[(curR2,curB2)] +=1
    ans = 0
    for key,i in d1.items():
        cur = i*d2[(key[1][::-1],key[0][::-1])]
        ans +=cur
    print(ans)
    return

import sys,copy,bisect,itertools,heapq,math
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,inf
mod = 10**9 + 7
inf = 10**18

if __name__ == '__main__':
    examB()
