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;
template <class T>
inline T MAX(T a, T b) {
  if (a > b) return a;
  return b;
}
template <class T>
inline T MIN(T a, T b) {
  if (a < b) return a;
  return b;
}
template <class T>
inline T ABS(T x) {
  if (x < 0) return -x;
  return x;
}
inline void OPEN(const string &s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
const static int inf = 1000000000;
int n, t, c, x;
int arr[200005];
int idx = 0;
int main() {
  scanf("%d%d%d", &n, &t, &c);
  for (int(i) = (0); (i) < (n); ++(i)) {
    scanf("%d", &arr[i]);
  }
  long long res = 0;
  int nowSz = 0, awalIdx = 0;
  for (int i = 0; i < c; i++) {
    if (arr[i] > t) {
      nowSz = 0;
      awalIdx = -1;
    } else {
      nowSz++;
      awalIdx = i;
    }
  }
  if (nowSz == c) res++;
  for (int i = c; i < n; i++) {
    if (arr[i] > t) {
      nowSz = 0;
    } else {
      nowSz = min(nowSz + 1, c);
    }
    if (nowSz == c) res++;
  }
  printf("%I64d\n", res);
  return 0;
}
