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 D {

	public static void main(String[] args) {
        FastScanner in = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int t = 1;//in.nextInt(), tt = 0;
        while(t-->0) { 
            int n = in.nextInt(), k = in.nextInt();
            char ch[] = new char[k*k];
            char a[] = new char[k];
            for(char i='a';i-'a'<k;i++) a[i-'a'] = i;      
            char c = 'a'; ch[0] = 'a';
            int ind = 0;
            for(int i=1;i<k*k;i++){
            	if(ch[i-1]==c){
            		ch[i] = a[ind++];
            		if(ind==k){
            			c++; ind = c-'a';
            		}
				}
				else ch[i] = c;
            }
            int mod = k*k;
            for(int i=0;i<n;i++) out.print(ch[i%mod]);
        }
        out.flush();
	}
	
	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		
		String next() {
			while(!st.hasMoreTokens())
				try { st = new StringTokenizer(br.readLine()); }
				catch(IOException e) {}
			return st.nextToken();
		}
		
		String nextLine(){
			try{ return br.readLine(); } 
			catch(IOException e) { } return "";
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		
		int[] readArray(int n) {
			int a[] = new int[n];
			for(int i=0;i<n;i++) a[i] = nextInt();
			return a;
		}
	}

	static final Random random = new Random();

	static void ruffleSort(int[] a){
		int n = a.length;
		for(int i=0;i<n;i++){
			int j = random.nextInt(n), temp = a[j];
			a[j] = a[i]; a[i] = temp;
		}
		Arrays.sort(a); 	
	}
}