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;
const int N = 1000005;
int f[N][20], n, t, c, ans;
inline void fr(int &num) {
  num = 0;
  char c = getchar();
  int p = 1;
  while (c < '0' || c > '9') c == '-' ? p = -1, c = getchar() : c = getchar();
  while (c >= '0' && c <= '9') num = num * 10 + c - '0', c = getchar();
  num *= p;
}
inline int Max(register int p, register int q) { return p > q ? p : q; }
int main() {
  fr(n), fr(c), fr(t);
  for (register int i = 1; i <= n; ++i) fr(f[i][0]);
  for (register int j = 1; (1 << j - 1) <= n; ++j)
    for (register int i = 1; i + (1 << j) - 1 <= n; ++i)
      f[i][j] = Max(f[i][j - 1], f[i + (1 << j - 1)][j - 1]);
  int k = log2(t);
  for (register int i = 1; i + t - 1 <= n; ++i)
    ans += (Max(f[i][k], f[i + t - (1 << k)][k]) <= c ? 1 : 0);
  printf("%d", ans);
  return 0;
}
