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
// Main Code at the Bottom
import java.util.*;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.io.*; 
public class Main{
	//Fast IO class
    static class FastReader {
        BufferedReader br; 
        StringTokenizer st; 
        public FastReader() {
        	boolean env=System.getProperty("ONLINE_JUDGE") != null;
        	if(!env) {
        		try {
					br=new BufferedReader(new FileReader("input.txt"));
				} catch (FileNotFoundException e) {
					e.printStackTrace();
				}
        	}
        	else 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; 
        } 
    }     
    static long MOD=(long)1e9+7;
    //debug
    static void debug(Object... o) {
        System.out.println(Arrays.deepToString(o));
    }
    static FastReader sc=new FastReader();
    static PrintWriter out=new PrintWriter(System.out);  
    //Global variables and functions
    static int find(String s) {
    	int ans=0;
    	for(int i=0;i<s.length()-1;i++) {
    		for(int j=i+1;j<s.length()-1;j++) {
    			if(s.charAt(i)==s.charAt(j) && s.charAt(i+1)==s.charAt(j+1)) ans++;
    		}
    	}
    	return ans;
    }
    //Main function(The main code starts from here)
    public static void main (String[] args) throws java.lang.Exception {
    	int test=1;
    	//test=sc.nextInt();
    	while(test-->0) {
    		int n=sc.nextInt(),k=sc.nextInt();
    		StringBuilder sb=new StringBuilder();
    		sb.append("a");
    		for(char i='a';i<(char)(k+'a'-1);i++) {
    			for(char j=(char)(k+'a'-1);j>(i+1);j--) {
    				sb.append(j);
    				sb.append(i);
    			}
    			sb.append((char)(i+1));
    			sb.append((char)(i+1));
    		}
    		for(char i=(char)(k+'a'-2);i>='a';i--) sb.append(i);
    		for(int i=0;i<n;i++) out.print(sb.charAt(i%(k*k)));
    		
    	}
        out.flush();
        out.close();
    }
}