The prison of your city has n prisoners. As the prison can't accommodate all of them, the city mayor has decided to transfer c of the prisoners to a prison located in another city.

For this reason, he made the n prisoners to stand in a line, with a number written on their chests. The number is the severity of the crime he/she has committed. The greater the number, the more severe his/her crime was.

Then, the mayor told you to choose the c prisoners, who will be transferred to the other prison. He also imposed two conditions. They are,

  * The chosen c prisoners has to form a contiguous segment of prisoners. 
  * Any of the chosen prisoner's crime level should not be greater then t. Because, that will make the prisoner a severe criminal and the mayor doesn't want to take the risk of his running away during the transfer. 



Find the number of ways you can choose the c prisoners.

Input

The first line of input will contain three space separated integers n (1 ≤ n ≤ 2·105), t (0 ≤ t ≤ 109) and c (1 ≤ c ≤ n). The next line will contain n space separated integers, the ith integer is the severity ith prisoner's crime. The value of crime severities will be non-negative and will not exceed 109. 

Output

Print a single integer — the number of ways you can choose the c prisoners.

Examples

Input

4 3 3
2 3 1 1


Output

2


Input

1 1 1
2


Output

0


Input

11 4 2
2 2 0 7 3 2 2 4 9 1 4


Output

6
import sys

def pyes_no(condition) :
  if condition :
    print "YES"
  else :
    print "NO"

def plist(a, s = ' ') :
  print s.join(map(str, a))

def rint() :
  return int(sys.stdin.readline())

def rints() :
  return map(int, sys.stdin.readline().split())

def rfield(n, m = None) :
  if m == None :
    m = n
  
  field = []
  for i in xrange(n) :
    chars = sys.stdin.readline().strip()
    assert(len(chars) == m)
    field.append(chars)
  return field

def digits(x, p) :
  digits = []
  while x > 0 :
    digits.append(x % p)
    x /= p
  return digits

def modpower(a, n, mod) :
  r = a ** (n % 2)
  if n > 1 :
    r *= modpower(a, n / 2, mod) ** 2
  return r % mod

def gcd(a, b) :
  if a > b :
    a, b = b, a
  
  while a > 0 :
    a, b = b % a, a

  return b

n, t, c = rints()
a = rints()

bad = sum([x > t for x in a[:c]])
count = bad == 0
for pop, push in zip(a[:n - c], a[c:]) :
  bad -= pop > t
  bad += push > t
  count += bad == 0

print int(count)
