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:
# -*- coding: utf-8 -*-

import sys
from fractions import gcd

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = 10 ** 19
MOD = 10 ** 9 + 7

def bisearch_min(mn, mx, func):
    """ 条件を満たす最小値を見つける二分探索 """

    ok = mx
    ng = mn
    while ng+1 < ok:
        mid = (ok+ng) // 2
        if func(mid):
            # 下を探しに行く
            ok = mid
        else:
            # 上を探しに行く
            ng = mid
    return ok

def bisearch_max(mn, mx, func):
    """ 条件を満たす最大値を見つける二分探索 """

    ok = mn
    ng = mx
    while ok+1 < ng:
        mid = (ok+ng) // 2
        if func(mid):
            # 上を探しに行く
            ok = mid
        else:
            # 下を探しに行く
            ng = mid
    return ok

for _ in range(INT()):
    a, b, c, d = MAP()

    # そもそも無理
    if a < b or d < b:
        No()
        continue
    
    # b個減ってd個増える繰り返しなので、このgより細かく値が動くことはない
    g = gcd(b, d)

    # cより大きくてb未満が存在するかどうか
    mn = bisearch_min(-INF, INF, lambda m: a%g + g*m > c)
    mx = bisearch_max(-INF, INF, lambda m: a%g + g*m < b)
    if mn > mx:
        Yes()
    else:
        No()