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
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Collections;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Stack;
import java.util.StringTokenizer;


public class Problem2 {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int t = Integer.parseInt(st.nextToken());
        int c = Integer.parseInt(st.nextToken());
        PriorityQueue<Integer> b = new PriorityQueue<Integer>(10, Collections.reverseOrder());
        st = new StringTokenizer(br.readLine());
        int y = 0;
        while(y<n&&b.size()<c) {
            b.add(Integer.parseInt(st.nextToken()));
            if(b.peek()>t) {
                b.clear();
            }
            y++;
        }
        int ans = 0;
        for (int i = y; i < n; ) {
            if(b.remove()<=t) {
                ans++;
                if(st.hasMoreTokens()) {
                    b.add(Integer.parseInt(st.nextToken()));
                    i++;
                }
            }
                else {
                    b.clear();
                    int x = i;
                    while(x<n&&b.size()<c) {
                        b.add(Integer.parseInt(st.nextToken()));
                        if(b.peek()>t) {
                            b.clear();
                        }
                        x++;
                    }
                    i = x;
                }
                
            }
        if(b.size()==c&&b.remove()<=t) {
            ans++;
        }
        System.out.println(ans);
    }
}
