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>
int l[200001];
int main() {
  long long i, j, n, t, c, p, m;
  while (scanf("%I64d%I64d%I64d", &n, &t, &c) != EOF) {
    p = 0;
    m = -1;
    for (i = 1; i <= n; i++) scanf("%d", &l[i]);
    for (i = 1; i <= n;) {
      if (m != 0) {
        if (i + c - 1 > n) break;
        for (j = 0; j < c; j++)
          if (l[i + j] > t) break;
        if (j == c) {
          p++;
          m = 0;
          i = i + c;
        } else
          i = i + j + 1;
      } else {
        if (l[i] <= t) {
          p++;
          i++;
        } else {
          i++;
          m = -1;
        }
      }
    }
    printf("%d\n", p);
  }
}
