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
{
	PrintWriter out = new PrintWriter(System.out);
	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer tok = new StringTokenizer("");
    String next() throws IOException {
        if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
        return tok.nextToken();
    }
    int ni() throws IOException { return Integer.parseInt(next()); }
    long nl() throws IOException { return Long.parseLong(next()); }
    
    long mod=1000000007;
    
    void solve() throws IOException {
        int n=ni(),k=ni();
        int[]A=new int[n];
        int[]B=new int[k];
        
        if (k==1) {
            for (int i=0;i<n;i++) out.print("a");
            out.println();
            out.flush();
            return;
        }
        
        B[k-1]=1;
        int p=1;
        while (p<n) {
            A[p]=B[A[p-1]];
            B[A[p-1]]++;
            if (A[p]==k-1 && A[p-1]==k-1) { p++; break; }
            p++;
        }
        int q=0;
        while (p<n) {
            A[p]=A[q];
            p++;
            q++;
        }
        for (int i=0;i<n;i++) {
            char c=(char)(A[i]);
            c+=97;
            out.print(c);
        }
        out.println();
        out.flush();
    }
    
    int len(long p) {
        int r=0;
        while (p>0) { r++; p/=10; }
        return r;
    }
    
    int gcd(int a,int b) { return(b==0?a:gcd(b,a%b)); }
    long gcd(long a,long b) { return(b==0?a:gcd(b,a%b)); }
    long mp(long a,long p) { long r=1; while(p>0) { if ((p&1)==1) r=(r*a)%mod; p>>=1; a=(a*a)%mod; } return r; }
    
    public static void main(String[] args) throws IOException {
        new Main().solve();
    }
}