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.util.Scanner;
import java.util.TreeMap;

public class B {

	static int N,T,C;
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		N=sc.nextInt();
		T=sc.nextInt();
		C=sc.nextInt();
		int[]a = new int[N];
		TreeMap<Integer,Integer>map=new TreeMap<Integer,Integer>();
		for(int i=0;i<N;i++){
			a[i]=sc.nextInt();
		}
		for(int i=0;i<C;i++){
			if(map.containsKey(a[i]))
				map.put(a[i], map.get(a[i])+1);
			else{
				map.put(a[i], 1);				
			}

		}
		int ans=0;
		if(map.lastKey()<=T){
			ans++;
		}
		for(int i=C;i<N;i++){
			//delete
			int idx=i-C;
			if(map.get(a[idx])>=2){
				map.put(a[idx], map.get(a[idx])-1);
			}
			else{
				map.remove(a[idx]);
			}
			
			if(map.containsKey(a[i]))
				map.put(a[i], map.get(a[i])+1);
			else
				map.put(a[i], 1);
			
			if(map.lastKey()<=T){
				ans++;
			}
			

		}
		System.out.println(ans);
	}
}
