Let's define the cost of a string s as the number of index pairs i and j (1 ≤ i < j < |s|) such that s_i = s_j and s_{i+1} = s_{j+1}.

You are given two positive integers n and k. Among all strings with length n that contain only the first k characters of the Latin alphabet, find a string with minimum possible cost. If there are multiple such strings with minimum cost — find any of them.

Input

The only line contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 26).

Output

Print the string s such that it consists of n characters, each its character is one of the k first Latin letters, and it has the minimum possible cost among all these strings. If there are multiple such strings — print any of them.

Examples

Input


9 4


Output


aabacadbb


Input


5 1


Output


aaaaa

Input


10 26


Output


codeforces
import java.io.*;
import java.util.*;

public class D {
    public static void main(String args[]) throws Exception{
        BufferedReader br = new BufferedReader( new InputStreamReader(System.in));
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int N = sc.nextInt();
        int K = sc.nextInt();
        
        char ch[] = new char[K];

        for(char c ='a'; c<'a'+K; c++){
            ch[c-'a'] = c;
        }

        String pattern = "";

        for(int i=0; i<K; i++){
            int j = 0;
            for(int k=i; k<K; k++){
                if(k!=K-1)
                pattern =pattern+""+ch[k]+""+ch[i];

                else
                pattern=pattern+""+ch[k];
            }
        }

        if(N<=pattern.length()){
            out.println(pattern.substring(0, N));
        }else{
            StringBuilder str = new StringBuilder();
            
            int len = pattern.length();
            int rem = N%len;
            int div = N/len;
            String ans = "";
            for(int i=0; i<div; i++){
                // ans+=""+pattern;
                str.append(pattern);
            }
            str.append(pattern.substring(0,rem));
            out.println(str);

        }
        out.flush();
        // System.out.println(pattern);
    }    
}
