Alexey, a merry Berland entrant, got sick of the gray reality and he zealously wants to go to university. There are a lot of universities nowadays, so Alexey is getting lost in the diversity — he has not yet decided what profession he wants to get. At school, he had bad grades in all subjects, and it's only thanks to wealthy parents that he was able to obtain the graduation certificate.

The situation is complicated by the fact that each high education institution has the determined amount of voluntary donations, paid by the new students for admission — n_{i} berubleys. He cannot pay more than n_{i}, because then the difference between the paid amount and n_{i} can be regarded as a bribe!

Each rector is wearing the distinctive uniform of his university. Therefore, the uniform's pockets cannot contain coins of denomination more than r_{i}. The rector also does not carry coins of denomination less than l_{i} in his pocket — because if everyone pays him with so small coins, they gather a lot of weight and the pocket tears. Therefore, a donation can be paid only by coins of denomination x berubleys, where l_{i} ≤ x ≤ r_{i} (Berland uses coins of any positive integer denomination). Alexey can use the coins of different denominations and he can use the coins of the same denomination any number of times. When Alexey was first confronted with such orders, he was puzzled because it turned out that not all universities can accept him! Alexey is very afraid of going into the army (even though he had long wanted to get the green uniform, but his dad says that the army bullies will beat his son and he cannot pay to ensure the boy's safety). So, Alexey wants to know for sure which universities he can enter so that he could quickly choose his alma mater.

Thanks to the parents, Alexey is not limited in money and we can assume that he has an unlimited number of coins of each type.

In other words, you are given t requests, each of them contains numbers n_{i}, l_{i}, r_{i}. For each query you need to answer, whether it is possible to gather the sum of exactly n_{i} berubleys using only coins with an integer denomination from l_{i} to r_{i} berubleys. You can use coins of different denominations. Coins of each denomination can be used any number of times.


-----Input-----

The first line contains the number of universities t, (1 ≤ t ≤ 1000) Each of the next t lines contain three space-separated integers: n_{i}, l_{i}, r_{i} (1 ≤ n_{i}, l_{i}, r_{i} ≤ 10^9; l_{i} ≤ r_{i}).


-----Output-----

For each query print on a single line: either "Yes", if Alexey can enter the university, or "No" otherwise.


-----Examples-----
Input
2
5 2 3
6 4 5

Output
Yes
No



-----Note-----

You can pay the donation to the first university with two coins: one of denomination 2 and one of denomination 3 berubleys. The donation to the second university cannot be paid.

Difficulty: MEDIUM

Solutions:
t = int(input())
for _ in range(t):
	(n, l, r) = map(int, input().split())
	print('No' if n // l * r < n else 'Yes')


import sys
sys.setrecursionlimit(1500)
inst_num = int(input())
n = 0
l = 0
r = 0

def payment_possible(total, l, r):
	max_coins = int(total / l)
	if r * max_coins >= total:
		return True
	else:
		return False
	if total < coins[0]:
		return False
	return False
for i in range(0, inst_num):
	line_data = input().split()
	n = int(line_data[0])
	l = int(line_data[1])
	r = int(line_data[2])
	if n > 0 and l == 1:
		print('Yes')
		continue
	if n >= l and n <= r:
		print('Yes')
		continue
	if payment_possible(n, l, r):
		print('Yes')
	else:
		print('No')


def possible(numbers):
	return int(numbers[0]) // int(numbers[1]) != 0 and int(numbers[0]) % int(numbers[1]) // (int(numbers[0]) // int(numbers[1])) <= int(numbers[2]) - int(numbers[1]) and (int(numbers[0]) % int(numbers[1]) % int(numbers[0]) // int(numbers[1]) + int(numbers[0]) % int(numbers[1]) % (int(numbers[0]) // int(numbers[1])) <= (int(numbers[2]) - int(numbers[1])) * (int(numbers[0]) // int(numbers[1]) - int(numbers[0]) // int(numbers[2])))
for i in range(0, int(input())):
	print('Yes' if possible(input().split(' ')) else 'No')


class CodeforcesTask397BSolution:

	def __init__(self):
		self.result = ''
		self.t = 0
		self.queries = []

	def read_input(self):
		self.t = int(input())
		for _ in range(self.t):
			self.queries.append([int(x) for x in input().split(' ')])

	def process_task(self):
		res = []
		for query in self.queries:
			k = query[0] // query[1]
			res.append('Yes' if k * query[2] >= query[0] else 'No')
		self.result = '\n'.join(res)

	def get_result(self):
		return self.result
Solution = CodeforcesTask397BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())


T = int(input())
while T > 0:
	T -= 1
	(n, l, r) = list(map(int, input().split()))
	t = n // l
	if t * r >= n:
		print('Yes')
	else:
		print('No')


for i in range(int(input())):
	(n, l, r) = map(int, input().split())
	print('No' if n // l * r < n else 'Yes')


t = int(input())
for i in range(0, t):
	(n, l, r) = [int(s) for s in input().split()]
	print('Yes' if n % l <= (r - l) * (n // l) else 'No')


t = int(input())
while t > 0:
	t -= 1
	(n, l, r) = map(int, input().split())
	k = n // l
	if r * k >= n:
		print('Yes')
	else:
		print('No')


import sys
sys.setrecursionlimit(1000000)

def solve():
	(tests,) = rv()
	for test in range(tests):
		(n, l, r) = rv()
		largestused = (n + r - 1) // r
		totalnum = largestused * r
		mostcansubtract = largestused * (r - l)
		lowerbound = totalnum - mostcansubtract
		if lowerbound <= n:
			print('Yes')
		else:
			print('No')

def rv():
	return list(map(int, input().split()))

def rl(n):
	return [list(map(int, input().split())) for _ in range(n)]
if sys.hexversion == 50594544:
	sys.stdin = open('test.txt')
solve()