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;
long long int MOD = 1000000007;
template <typename T>
T gcd(T x, T y) {
  return y == 0 ? x : gcd(y, x % y);
}
template <typename T>
T lcm(const T &a, const T &b) {
  return (a * b) / gcd(a, b);
}
long long int nextPowerOf2(long long int n) {
  long long int count = 0;
  if (n && !(n & (n - 1))) return n;
  while (n != 0) {
    n >>= 1;
    count += 1;
  }
  return 1 << count;
}
long long int raiseToPow(long long int x, long long int power) {
  long long int result;
  long long int i;
  result = 1;
  for (i = 1; i <= power; i++) {
    result = result * x;
  }
  return (result);
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cout.precision(12);
  cout << fixed;
  int n, t, c;
  cin >> n >> t >> c;
  vector<long long int> arr(n), pos;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] > t) pos.push_back(i);
  }
  long long int ans = 0;
  if (pos.size() == 0) {
    cout << n - c + 1;
    return 0;
  }
  if (pos[0] >= c) ans += pos[0] - c + 1;
  for (int i = 1; i < pos.size(); i++) {
    if (pos[i] - pos[i - 1] >= c) ans += pos[i] - pos[i - 1] - c;
  }
  if (n - pos[pos.size() - 1] >= c) ans += n - pos[pos.size() - 1] - c;
  cout << ans;
  cerr << "\nTime elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s.\n";
  return 0;
}
