There are n programmers that you want to split into several non-empty teams. The skill of the i-th programmer is a_i. You want to assemble the maximum number of teams from them. There is a restriction for each team: the number of programmers in the team multiplied by the minimum skill among all programmers in the team must be at least x.

Each programmer should belong to at most one team. Some programmers may be left without a team.

Calculate the maximum number of teams that you can assemble.

Input

The first line contains the integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first line of each test case contains two integers n and x (1 ≤ n ≤ 10^5; 1 ≤ x ≤ 10^9) — the number of programmers and the restriction of team skill respectively.

The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9), where a_i is the skill of the i-th programmer.

The sum of n over all inputs does not exceed 10^5.

Output

For each test case print one integer — the maximum number of teams that you can assemble. 

Example

Input


3
5 10
7 11 2 9 5
4 8
2 4 2 3
4 11
1 3 3 7


Output


2
1
0
 
# from math import factorial as fac
from collections import defaultdict
# from copy import deepcopy
import sys, math
f = None
try:
	f = open('q1.input', 'r')
except IOError:
	f = sys.stdin
if 'xrange' in dir(__builtins__):
	range = xrange
# print(f.readline())
sys.setrecursionlimit(10**2)

	
def print_case_iterable(case_num, iterable):
	print("Case #{}: {}".format(case_num," ".join(map(str,iterable))))

def print_case_number(case_num, iterable):
	print("Case #{}: {}".format(case_num,iterable))

def print_iterable(A):
	print (' '.join(A))

def read_int():
	return int(f.readline().strip())
def read_int_array():
	return [int(x) for x in f.readline().strip().split(" ")]
def rns():
	a =  [x for x in f.readline().split(" ")]
	return int(a[0]), a[1].strip()
def read_string():
	return list(f.readline().strip())
def ri():
	return int(f.readline().strip())
def ria():
	return [int(x) for x in f.readline().strip().split(" ")]
def rns():
	a =  [x for x in f.readline().split(" ")]
	return int(a[0]), a[1].strip()
def rs():
	return list(f.readline().strip())
def bi(x):
	return bin(x)[2:]





from collections import deque
import math
NUMBER = 10**9 + 7
# NUMBER = 998244353
def factorial(n) : 
	M = NUMBER
	f = 1
  
	for i in range(1, n + 1):  
		f = (f * i) % M # Now f never can  
						# exceed 10^9+7  
	return f
def mult(a,b):
	return (a * b) % NUMBER

def minus(a , b):
	return (a - b) % NUMBER

def plus(a , b):
	return (a + b) % NUMBER

def egcd(a, b):
	if a == 0:
		return (b, 0, 1)
	else:
		g, y, x = egcd(b % a, a)
		return (g, x - (b // a) * y, y)

def modinv(a):
	m = NUMBER
	g, x, y = egcd(a, m)
	if g != 1:
		raise Exception('modular inverse does not exist')
	else:
		return x % m
def choose(n,k):
	if n < k:
		assert false
	return  mult(factorial(n), modinv(mult(factorial(k),factorial(n-k)))) % NUMBER
from collections import deque, defaultdict 
import heapq


from types import GeneratorType
def bootstrap(f, stack=[]):
	def wrappedfunc(*args, **kwargs):
		if stack:
			return f(*args, **kwargs)
		else:
			to = f(*args, **kwargs)
			while True:
				if type(to) is GeneratorType:
					stack.append(to)
					to = next(to)
				else:
					stack.pop()
					if not stack:
						break
					to = stack[-1].send(to)
			return to

	return wrappedfunc


def dfs(g, timeIn, timeOut,depths,parents): 
	# assign In-time to node u 
	cnt = 0
	# node, neig_i, parent, depth
	stack = [[1,0,0,0]]
	while stack:
		v,neig_i,parent,depth = stack[-1]
		parents[v] = parent
		depths[v] = depth
		# print (v)
		if neig_i == 0:
			timeIn[v] = cnt    
			cnt+=1
		while neig_i < len(g[v]):
			u = g[v][neig_i]
			if u == parent:
				neig_i+=1
				continue
			stack[-1][1] = neig_i + 1
			stack.append([u,0,v,depth+1])
			break
		if neig_i == len(g[v]):
			stack.pop()
			timeOut[v] = cnt
			cnt += 1
  
# def isAncestor(u: int, v: int, timeIn: list, timeOut: list) -> str: 
# 	return timeIn[u] <= timeIn[v] and timeOut[v] <= timeOut[u] 

cnt = 0
@bootstrap
def dfs(v,adj,timeIn, timeOut,depths,parents,parent=0,depth=0):
	global cnt
	parents[v] = parent
	depths[v] = depth
	timeIn[v] = cnt
	cnt+=1
	for u in adj[v]:
		if u == parent:
			continue
		yield dfs(u,adj,timeIn,timeOut,depths,parents,v,depth+1)
	timeOut[v] = cnt
	cnt+=1
	yield 

def gcd(a,b): 
	if a == 0: 
		return b 
	return gcd(b % a, a) 
  
# Function to return LCM of two numbers 
def lcm(a,b): 
	return (a*b) / gcd(a,b) 
def get_num_2_5(n):
	twos = 0
	fives = 0
	while n>0 and n%2 == 0:
		n//=2
		twos+=1
	while n>0 and n%5 == 0:
		n//=5
		fives+=1
	return (twos,fives)
def shift(a,i,num):
	for _ in range(num):
		a[i],a[i+1],a[i+2] = a[i+2],a[i],a[i+1]

def equal(x,y):
	return abs(x-y) <= 1e-9
# def leq(x,y):
# 	return x-y <= 1e-9
def getAngle(a, b, c):
	ang = math.degrees(math.atan2(c[1]-b[1], c[0]-b[0]) - math.atan2(a[1]-b[1], a[0]-b[0]))
	return ang + 360 if ang < 0 else ang
def getLength(a,b):
	return math.sqrt((a[0]-b[0])**2 + (a[1]-b[1])**2)


from heapq import heapify, heappush, heappop

def solution(a,n,x):
	for i in range(n):
		a[i] = (x+a[i]-1)//a[i]
	d = defaultdict(int)
	for y in a:
		d[y]+=1
	nums = sorted(set(a))
	res = 0
	for i in range(len(nums)):
		res+=d[nums[i]]//nums[i]
		if i <len(nums)-1:
			d[nums[i+1]]+=d[nums[i]]%nums[i]
	return res



	

	

	
def main():
	T = 1
	T = ri()
	for i in range(T):
		# n = ri()
		# s = rs()
		n,x	= ria()
		a = ria()

		x = solution(a,n,x)
		# continue
		if 'xrange' not in dir(__builtins__):
			print(x)
		else:
			print >>output,str(x)# "Case #"+str(i+1)+':',
	if 'xrange' in dir(__builtins__):
		print(output.getvalue())
		output.close()

if 'xrange' in dir(__builtins__):
	import cStringIO
	output = cStringIO.StringIO()
#example usage:
#    for l in res:
#       print >>output, str(len(l)) + ' ' +  ' '.join(l)

if __name__ == '__main__':
	main()
