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>
long long R = 7 + 1e9, NUMTESTCASE;
const int NN = 10 + 1e5;
const double pi = acos(-1.0);
int di[4] = {1, 0, -1, 0}, dj[4] = {0, 1, 0, -1},
    DI[8] = {1, 1, 0, -1, -1, -1, 0, 1}, DJ[8] = {0, 1, 1, 1, 0, -1, -1, -1};
using namespace std;
int main() {
  int x, t, n, mx, k;
  cin >> n >> mx >> k;
  vector<int> a[n];
  t = 0;
  for (int i = (0); i <= (n - 1); ++i) {
    scanf("%d", &x);
    if (x <= mx)
      a[t].push_back(1);
    else
      t++;
  }
  int ans = 0;
  for (int i = (0); i <= (n - 1); ++i) ans += max(0, (int)a[i].size() - k + 1);
  cout << ans;
  return 0;
}
