You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.

You have to answer q independent test cases.

Input

The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.

The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.

Output

For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.

You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).

Example

Input


4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18


Output


YES
NO
NO
YES
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
# sys.setrecursionlimit(int(pow(10, 2)))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]


# @lru_cache(None)




for _ in range(l()[0]):
    a,b,n,s=l()
    x=min(a,s//n)
    y=(s-x*n)
    if(y<=b):
        print("YES")
    else:
        print("NO")