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 ma[200010 << 2];
void init() { memset(ma, 0, sizeof(ma)); }
void update(int l, int r, int u, int pos, int val) {
  if (l == r) {
    ma[u] = val;
    return;
  }
  int mid = (l + r) / 2;
  if (pos <= mid)
    update(l, mid, u << 1, pos, val);
  else
    update(mid + 1, r, u << 1 | 1, pos, val);
  ma[u] = max(ma[u << 1], ma[u << 1 | 1]);
}
int query(int l, int r, int L, int R, int u) {
  if (l == L && r == R) return ma[u];
  int mid = (l + r) / 2;
  if (R <= mid)
    return query(l, mid, L, R, u << 1);
  else if (L > mid)
    return query(mid + 1, r, L, R, u << 1 | 1);
  else
    return max(query(l, mid, L, mid, u << 1),
               query(mid + 1, r, mid + 1, R, u << 1 | 1));
}
int main() {
  int n, t, c;
  while (~scanf("%d", &n)) {
    init();
    scanf("%d%d", &t, &c);
    for (int i = 0; i < n; ++i) {
      int tmp;
      scanf("%d", &tmp);
      update(1, n, 1, i + 1, tmp);
    }
    int ans = 0;
    for (int i = 1; i <= n && i + c - 1 <= n; ++i) {
      if (query(1, n, i, i + c - 1, 1) <= t) ++ans;
    }
    printf("%d\n", ans);
  }
  return 0;
}
