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.lang.*;
import java.io.*;
public class Main
{
    static FastReader sc=new FastReader(); 
	public static void main (String[] args) throws java.lang.Exception
	{
        PrintWriter w=new PrintWriter(System.out);
		int tc=1;
		while(tc-->0){
		    int n=sc.nextInt();
		    int k=sc.nextInt();
		    if(k==1){
		        String ans="";
		        for(int i=0;i<n;i++)w.write("a");
		        w.write("\n");
		    }
		    
		    else{
		        String s="";
    		    int[][] a=new int[k][k];
    		    for(int i=0;i<k;i++){
    		        s+=(char)('a'+i);
    		        s+=(char)('a'+i);
    		        a[i][i]=1;
    		        a[i][(i+1)%k]=1;
    		    }
    		    s+='a';
    		    int p=0;
    		    while(true){
    		        int f=0;
    		        for(int i=0;i<k;i++){
        		        if(a[p][i]==0)
        		        {a[p][i]=1;p=i;s+=(char)('a'+i);f=1;}
        		    }
        		    if(f==0)break;
    		    }
    		    int slen=s.length();
    		    
		        if(n<=slen)w.write(s.substring(0,n)+"\n");
    		    else{
    		        s=s.substring(0,slen-1);
    		        String ans=s;
    		        n-=slen;
    		        n++;
    		        while(n>0){
    		            if(n>=slen-1){
    		                s+=ans;
    		                n-=slen;
    		                n++;
    		            }
    		            else{
    		                s+=ans.substring(0,n);
    		                n=0;
    		            }
    		        }
    		        w.write(s+"\n");
    		    }
		    }
		    
		  //  w.write("\n");
		}
		w.flush();
		w.close();
	}
	public static int coun(long n){
	    int c=0;
	    while(n>0){
	        c++;
	        n/=10;
	    }
	    return c;
	}
}
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; 
        } 
    }