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;
const int maxn = 200005;
int A[maxn], p[maxn][20];
int getmax(int l, int r) {
  int j = 0;
  while (1 << (j + 1) <= r - l + 1) ++j;
  return max(p[l][j], p[r - (1 << j) + 1][j]);
}
int main(int argc, char const *argv[]) {
  int n, T, c;
  cin >> n >> T >> c;
  for (int(i) = (int)(1); (i) <= (int)(n); ++(i)) cin >> A[i];
  for (int(i) = (int)(1); (i) <= (int)(n); ++(i)) p[i][0] = A[i];
  for (int j = 1; (1 << j) <= n; ++j)
    for (int i = 1; i + (1 << j) - 1 <= n; ++i)
      p[i][j] = max(p[i][j - 1], p[i + (1 << (j - 1))][j - 1]);
  int res = 0;
  for (int i = 1; i + c - 1 <= n; ++i) res += getmax(i, i + c - 1) <= T;
  printf("%d\n", res);
  return 0;
}
