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;
using namespace std;
long long gcd(long long x, long long y) { return (!y) ? x : gcd(y, x % y); }
long long lcm(double x, double y) { return ((x / gcd(x, y)) * y); }
long long fact(long long n) { return !n ? 1 : n * fact(n - 1); }
long long nCr(long long n, long long r) {
  long long ans = 1;
  long long div = 1;
  for (long long i = r + 1; i <= n; i++) {
    ans = ans * i;
    ans /= div;
    div++;
  }
  return ans;
}
void A_KHALIFA() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
bool is_prime(long long n) {
  if (n == 1 || n == 0) return 0;
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
long long arr[1000000];
int main() {
  A_KHALIFA();
  long long n, t, c, a;
  long long count = 0, sum = 0;
  cin >> n >> t >> c;
  for (int i = 0; i < n; i++) {
    cin >> a;
    if (a <= t) {
      count++;
    }
    if (a > t) count = 0;
    if (count >= c) sum++;
  }
  cout << sum << '\n';
}
