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
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using namespace std;
bool is_prime(int n) {
  for (int i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
inline long long getPow(long long a, long long b) {
  long long res = 1, tp = a;
  while (b) {
    if (b & 1ll) {
      res *= tp;
      res %= 1000000007;
    }
    tp *= tp;
    tp %= 1000000007;
    b >>= 1ll;
  }
  return res;
}
inline long long to_ll(string s) {
  long long cur = 0;
  for (int i = 0; i < s.size(); i++) {
    cur = cur * 10 + s[i] - '0';
  }
  return cur;
}
inline string to_str(long long x) {
  string s = "";
  while (x) {
    s += char(x % 10 + '0');
    x /= 10;
  }
  reverse(s.begin(), s.end());
  return s;
}
inline int nxt() {
  int x;
  scanf("%d", &x);
  return x;
}
void ok() {
  puts("YES");
  exit(0);
}
void panic() {
  puts("NO");
  exit(0);
}
int main() {
  long long n = nxt(), t = nxt(), c = nxt();
  long long tp = 0, ans = 0;
  for (int i = 0; i < n; i++) {
    long long cur = nxt();
    if (cur <= t) {
      tp++;
    } else
      tp = 0;
    if (tp >= c) {
      ans++;
    }
  }
  cout << ans;
  return 0;
}
