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 st[200010][18], lg[200010];
int ask(int l, int r) {
  int k = lg[r - l + 1];
  return max(st[l][k], st[r - (1 << k) + 1][k]);
}
int main() {
  int n, mx, len, x;
  scanf("%d%d%d", &n, &mx, &len);
  int pre = 0;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &st[i][0]);
  }
  for (int j = 1; j <= 17; j++)
    for (int i = 1; i + (1 << j) - 1 <= n; i++)
      st[i][j] = max(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
  lg[0] = 1;
  for (int i = 2; i <= n; i++)
    if (i & (i - 1))
      lg[i] = lg[i - 1];
    else
      lg[i] = lg[i - 1] + 1;
  for (int i = 1; i + len - 1 <= n; i++) {
    if (ask(i, i + len - 1) <= mx) ans++;
  }
  cout << ans << endl;
  return 0;
}
