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 comment(linker, "/STACK:16777216")
using namespace std;
inline long FastMax(long x, long y) {
  return (((y - x) >> (32 - 1)) & (x ^ y)) ^ y;
}
inline long FastMin(long x, long y) {
  return (((y - x) >> (32 - 1)) & (x ^ y)) ^ x;
}
long IR[] = {0, -1, 0, 1, -1, -1, 1, 1};
long IC[] = {1, 0, -1, 0, 1, -1, -1, 1};
long N, t, c;
long A[1000007 + 7];
int main(void) {
  long i, j, v, Icase, k = 0;
  scanf("%ld%ld%ld", &N, &t, &c);
  for (i = 1; i <= N; i++) {
    scanf("%ld", &A[i]);
    if (A[i] > t) A[i] = -1;
  }
  A[0] = A[N + 1] = -1;
  long ans = 0;
  for (i = 1; i <= N + 1;) {
    if (A[i] < 0) {
      i++;
      continue;
    }
    long l = 1;
    for (i++; i <= N + 1 and A[i] >= 0; i++) {
      l++;
    }
    ans += max(l - c + 1, 0L);
  }
  printf("%ld\n", ans);
  return 0;
}
