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 a[200005];
int n, t, c;
class tree {
 private:
  vector<int> st;
  int N;

 public:
  tree(int N) {
    this->N = n;
    st.assign(4 * N, 0);
    build(1, 0, N - 1);
    cout << endl;
  }
  void build(int id, int l, int r) {
    if (l == r) {
      st[id] = a[l];
      return;
    }
    int mid = (l + r) / 2;
    int left = 2 * id;
    int right = left + 1;
    build(left, l, mid);
    build(right, mid + 1, r);
    st[id] = max(st[left], st[right]);
  }
  int getValue(int id, int l, int r, int qs, int qe) {
    if (l == qs && r == qe) return st[id];
    int mid = (l + r) / 2;
    if (qs > mid) return getValue(2 * id + 1, mid + 1, r, qs, qe);
    if (qe <= mid) return getValue(2 * id, l, mid, qs, qe);
    return max(getValue(2 * id, l, mid, qs, mid),
               getValue(2 * id + 1, mid + 1, r, mid + 1, qe));
  }
  int getValue(int qs, int qe) { return getValue(1, 0, N - 1, qs, qe); }
};
int main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int count = 0;
  scanf("%d%d%d", &n, &t, &c);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  tree tr(n);
  for (int i = 0; i < n - c + 1; ++i) {
    if (tr.getValue(i, i + c - 1) <= t) count++;
  }
  cout << count << endl;
  return 0;
}
