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>
using namespace std;
int main() {
  long long n, maxi, k;
  cin >> n >> k >> maxi;
  long long arr[n];
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
  }
  if (maxi == 1) {
    int ct = 0;
    for (long long i = 0; i < n; i++) {
      if (arr[i] <= k) ct++;
    }
    cout << ct;
    return 0;
  }
  long long s, e, sum = 0, foo = 0;
  for (long long i = 0; i < n;) {
    s = 0, e = 0;
    if (arr[i] <= k) {
      s = i;
      i++;
      int foo = 0;
      while (arr[i] <= k && i <= n - 1) {
        e = i;
        i++;
        foo = 1;
      }
      if (abs(e - s) - maxi + 1 >= 0 && foo == 1) {
        sum += e - s - maxi + 2;
      }
    } else
      i++;
  }
  cout << sum;
}
