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 PYTHON):

Solution:
#!/usr/bin/python
# -*- coding: utf-8 -*-
import sys,collections,itertools,re,math,fractions,decimal,random,array,bisect,heapq

# decimal.getcontext().prec = 50
# sys.setrecursionlimit(10000)
# MOD = 10**9 + 7

def gcd(a, b):
    """greatest common divisor"""
    if b == 0:
        return a
    else:
        return gcd(b, a % b)

def solve(f):
    t = f.read_int()
    ans = []

    def solve2():
        a, b, c, d = f.read_int_list()

        g1 = gcd(gcd(a, b), d)

        if d < b: return 'No'
        if a < b: return 'No'
        if b <= c+g1: return 'Yes'
        g2 = gcd(b, d)

        lo = ((c+1)-a)%g2
        if lo == 0 or lo + (b-c-1) >= g2: return 'No'

        return 'Yes'

    for _ in xrange(t):
        ans.append(solve2())

    return "\n".join(ans)


class Reader(object):
    def __init__(self, filename=None):
        self.file = open(filename) if filename is not None else None
        self.case = 1

    def __readline(self):
        return self.file.next().strip() if self.file else raw_input()

    def next_case(self):
        self.file.next()
        self.case += 1

    def read_int(self): return int(self.__readline())
    def read_float(self): return float(self.__readline())
    def read_long(self): return long(self.__readline())
    def read_decimal(self): return decimal.Decimal(self.__readline())
    def read_str(self): return self.__readline()

    def read_int_list(self): return map(int, self.__readline().split())
    def read_float_list(self): return map(float, self.__readline().split())
    def read_long_list(self): return map(long, self.__readline().split())
    def read_decimal_list(self): return map(decimal.Decimal, self.__readline().split())
    def read_str_list(self): return self.__readline().split()

if __name__ == '__main__':
    filename = sys.argv[1] if len(sys.argv) > 1 else None
    f = Reader(filename)
    if f.file:
        while True:
            print "Case #%d\n"%f.case, solve(f)
            try:
                f.next_case()
            except StopIteration:
                break
    else:
        print solve(f)