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.util.*;
import java.io.*;
import java.math.*;
 

public final class Solution{
	

	public static void main(String[] args) throws IOException{
//		Scanner sc=new Scanner(System.in);
		FastReader sc=new FastReader();
		PrintWriter writer=new PrintWriter(System.out);
		int tc=1;
		while(tc-->0) {
			int n=sc.nextInt(), k=sc.nextInt();
			char[] c=new char[k];
			for(int i=0;i<k;i++) {
				c[i]=(char)((int)'a'+i);
			}
			String ans="";
			int i=0;
			while(n>0 && i<k) {
				char in=c[i];
				ans+=in;
				n--;
				for(int j=i+1;j<k && n>0;j++) {
					if(n>1) {ans+=in;ans+=c[j];n-=2;}
					else {ans+=in; n--;}
				}
				i++;
			}
			int l=ans.length();
			while(n>ans.length()) {
				ans+=ans;
				n-=l;
				l=ans.length();
			}
			ans+=ans.substring(0, n);
//			l=0;
//			while(n>0 && l<ans.length()) {
//				ans+=ans.charAt(l++);
//				n--;
//			}
			writer.println(ans);
		}
		writer.flush();
		writer.close();
  }
	
  static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }
   
}