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 segtree[1000000];
int v[200000];
int init(int root, int beg, int end) {
  if (beg == end) {
    segtree[root] = v[beg];
    return v[beg];
  }
  int left = 2 * root + 1, right = 2 * root + 2;
  int mid = (beg + end) / 2;
  segtree[root] = max(init(left, beg, mid), init(right, mid + 1, end));
  return segtree[root];
}
int query(int root, int beg, int end, int qbeg, int qend) {
  if (qend < beg || end < qbeg) return 0;
  if (qbeg <= beg && qend >= end) return segtree[root];
  int left = 2 * root + 1, right = 2 * root + 2;
  int mid = (beg + end) / 2;
  return max(query(left, beg, mid, qbeg, qend),
             query(right, mid + 1, end, qbeg, qend));
}
int main(void) {
  int n, t, c;
  int ans = 0, i;
  scanf("%d%d%d", &n, &t, &c);
  for (i = 0; i < n; i++) scanf("%d", &v[i]);
  init(0, 0, n - 1);
  for (i = 0; i + c <= n; i++) {
    if (query(0, 0, n - 1, i, i + c - 1) <= t) ans++;
  }
  printf("%d\n", ans);
}
