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
/*
bts songs to dance to:
I need U
Run
ON
Filter
I'm fine
 */
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.System.out;
import java.util.*;
import java.io.*;
import java.math.*;

public class x1511D
{
    static ArrayDeque<Integer>[] edges;
    public static void main(String hi[]) throws Exception
    {
        BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(infile.readLine());
        int N = Integer.parseInt(st.nextToken());
        int K = Integer.parseInt(st.nextToken());
        edges = new ArrayDeque[K];
        for(int a=0; a < K; a++)
        {
            edges[a] = new ArrayDeque<Integer>();
            for(int b=0; b < K; b++)
                edges[a].add(b);
        }
        ArrayList<Integer> path = new ArrayList<Integer>();
        dfs(0, path);
        StringBuilder sb = new StringBuilder("a");
        path.remove(0);
        for(int i=0; i < N-1; i++)
            sb.append((char)(path.get(i%path.size())+'a'));
        System.out.println(sb);
    }
    public static void dfs(int curr, ArrayList<Integer> path)
    {
        while(edges[curr].size() > 0)
        {
            int next = edges[curr].poll();
            dfs(next, path);
        }
        path.add(curr);
    }
}