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 SZ = 2 * 100005;
int arr[SZ];
int csum[4 * SZ];
void init(int root, int b, int e) {
  if (b == e) {
    csum[root] = arr[b];
    return;
  }
  int l = (root * 2);
  int r = (l + 1);
  int m = (b + e) / 2;
  init(l, b, m);
  init(r, m + 1, e);
  csum[root] = max(csum[l], csum[r]);
}
int query(int root, int b, int e, int i, int j) {
  if (i == b and j == e) return csum[root];
  int l = (root * 2);
  int r = l + 1;
  int m = (b + e) / 2;
  if (j <= m) {
    return query(l, b, m, i, j);
  } else if (i > m) {
    return query(r, m + 1, e, i, j);
  } else
    return max(query(l, b, m, i, m), query(r, m + 1, e, m + 1, j));
}
int main(int argc, const char *argv[]) {
  int N, T, C;
  while (scanf("%d %d %d", &N, &T, &C) == 3) {
    csum[0] = 0;
    for (int i = 1; i <= N; i++) {
      scanf("%d", &arr[i]);
    }
    init(1, 1, N);
    int res = 0;
    for (int i = 1; i + C - 1 <= N; i++) {
      if (query(1, 1, N, i, (i + C - 1)) <= T) res++;
    }
    printf("%d\n", res);
  }
  return 0;
}
