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 tree[1000005], arr[200005];
void build(int node, int b, int e) {
  if (b == e) {
    tree[node] = arr[b];
    return;
  }
  int mid = (b + e) / 2;
  int left = node * 2;
  int right = left + 1;
  build(left, b, mid);
  build(right, mid + 1, e);
  tree[node] = max(tree[left], tree[right]);
}
int query(int node, int b, int e, int i, int j) {
  if (i <= b && j >= e) return tree[node];
  if (i > e || j < b) return 0;
  int mid = (b + e) / 2;
  int left = node * 2;
  int right = left + 1;
  int p = query(left, b, mid, i, j);
  int q = query(right, mid + 1, e, i, j);
  return max(p, q);
}
int main() {
  int n, t, c;
  scanf("%d %d %d", &n, &t, &c);
  for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
  build(1, 1, n);
  long long ans = 0;
  for (int i = 1, j = i + c - 1; j <= n; j++, i++) {
    int p = query(1, 1, n, i, j);
    if (p <= t) ans++;
  }
  cout << ans << endl;
}
