Given a string s of length n and integer k (1 ≤ k ≤ n). The string s has a level x, if x is largest non-negative integer, such that it's possible to find in s:

  * x non-intersecting (non-overlapping) substrings of length k, 
  * all characters of these x substrings are the same (i.e. each substring contains only one distinct character and this character is the same for all the substrings). 



A substring is a sequence of consecutive (adjacent) characters, it is defined by two integers i and j (1 ≤ i ≤ j ≤ n), denoted as s[i ... j] = "s_{i}s_{i+1} ... s_{j}".

For example, if k = 2, then:

  * the string "aabb" has level 1 (you can select substring "aa"), 
  * the strings "zzzz" and "zzbzz" has level 2 (you can select two non-intersecting substrings "zz" in each of them), 
  * the strings "abed" and "aca" have level 0 (you can't find at least one substring of the length k=2 containing the only distinct character). 



Zuhair gave you the integer k and the string s of length n. You need to find x, the level of the string s.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5) — the length of the string and the value of k.

The second line contains the string s of length n consisting only of lowercase Latin letters.

Output

Print a single integer x — the level of the string.

Examples

Input


8 2
aaacaabb


Output


2


Input


2 1
ab


Output


1


Input


4 2
abab


Output


0

Note

In the first example, we can select 2 non-intersecting substrings consisting of letter 'a': "(aa)ac(aa)bb", so the level is 2.

In the second example, we can select either substring "a" or "b" to get the answer 1.
//package Round533;

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        String s = in.next();
        int[][] a = new int[n + 1][2];
        int m = 0;
        for (int i = 1; i <= n; i++) {
            int x = s.charAt(i - 1) - 'a' + 1;
            if (m > 0 && s.charAt(i - 2) == s.charAt(i - 1)) a[m][1]++;
            else {
                m++;
                a[m][0] = x;
                a[m][1] = 1;
            }
        }
//        for (int i = 0; i <= 1; i++) {
//            for (int j = 1; j <= m; j++)
//                System.out.print(a[j][i]+ " ");
//            System.out.println();
//        }
        int ans = 0;
        for (int i = 1; i <= 26; i++) {
            int x = 0;
            for (int j = 1; j <= m; j++)
                if (a[j][0] == i) x += a[j][1] / k;
            ans = Math.max(ans, x);
        }
        System.out.println(ans);
    }
}
