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>
const int oo = 0x3f3f3f3f, mod = 1e9 + 7;
using namespace std;
int dx[]{1, -1, 0, 0, 1, 1, -1, -1};
int dy[]{0, 0, 1, -1, 1, -1, 1, -1};
long long OO = (1LL * oo * oo);
void K_K() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long gcd(long long a, long long b) {
  return b == 0 ? abs(a) : gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long n, t, c;
vector<int> v;
long long fun() {
  int cnt = 0, cc = 0;
  for (int i = 0; i < c; i++) {
    if (v[i] > t) cc++;
  }
  if (!cc) cnt++;
  for (int i = c; i < n; i++) {
    if (v[i - c] > t && cc) cc--;
    if (v[i] <= t && !cc)
      cnt++;
    else if (v[i] > t)
      cc++;
  }
  return cnt;
}
int main() {
  K_K();
  cin >> n >> t >> c;
  v.resize(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  cout << fun();
}
