Problem: 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

Solution (in JAVA):

Solution:
 
import java.io.*;
import java.util.*;
 
 
public class NewClass {
            public static void main(String[] args)throws Exception{

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        
        int  t=1;
          while(t-->0){
              String []num = br.readLine().split(" ");
              int n = Integer.parseInt(num[0]);
              int k = Integer.parseInt(num[1]);
            String st = "abcdefghijklmnopqrstuvwxyz";
            String stt="";
            String ma="";
            boolean ch=false;
            if(k==1){
                if(n>100){
                    for(int i=0;i<100;i++){
                        ma+="a";
                    }
                    int z = n%ma.length();
                    int x = n/ma.length();
                    for(int i=0;i<x;i++){
                        stt+=ma;
                    }
                    stt+=ma.substring(0, z);
                    System.out.println(stt);
                    return;
                }
                else{
                    for(int i=0;i<n;i++){
                        ma+="a";
                    }
                    System.out.println(ma);
                return;
                }
                
                
            }
            for(int i=0;i<k;i++){
                
                String a=String.valueOf(st.charAt(i));
                ma+=a;
                for(int j=i+1;j<k;j++){
                    String b=String.valueOf(st.charAt(j));
                    if(ma.length()+1==n){
                        ma+=b;
                        ch=true;
                        break;
                    }
                    if(ma.length()==n){
                        ch=true;
                        break;
                }
                    ma+=a+String.valueOf(st.charAt(j));
                }
                if(ch){
                    break;
                }
            }
            if(ch){
                System.out.println(ma);
            }
            else{
                int c=0;
                if(n>100&&ma.length()<10){
                    for(int i=0;i<10;i++){
                        ma+=ma;
                    }
                    int z = n%ma.length();
                    int x = n/ma.length();
                    for(int i=0;i<x;i++){
                        stt+=ma;
                    }
                    stt+=ma.substring(0, z);
                    System.out.println(stt);
                    return;
                }
                {
                    int z = n%ma.length();
                    int x = n/ma.length();
                    for(int i=0;i<x;i++){
                        stt+=ma;
                    }
                    stt+=ma.substring(0, z);
                    System.out.println(stt);
                }
            }
            
            
        }    
    }
           
}