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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
    public static void main(String[] args) {
        CP sc = new CP();
        int n = sc.nextInt(), k = sc.nextInt();
        StringBuilder s = new StringBuilder();
        for (int i = 0; i < k; i++) {
            s.append((char) (97 + i));
            for (int j = i + 1; j < k; j++) {
                s.append((char) (97 + i));
                s.append((char) (97 + j));
            }
        }
        int l = s.length();
        int d = n / l;
        int r = n % l;
        StringBuilder ans = new StringBuilder();
        for (int i = 0; i < d; i++)
            ans.append(s);
        ans.append(s.substring(0, r));
        System.out.println(ans);
    }

    /*****************************************************************************/
    static class CP {
        BufferedReader bufferedReader;
        StringTokenizer stringTokenizer;

        public CP() {
            bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        }

        int nextInt() {
            return Integer.parseInt(NNNN());
        }

        long nextLong() {
            return Long.parseLong(NNNN());
        }

        double nextDouble() {
            return Double.parseDouble(NNNN());
        }

        String NNNN() {
            while (stringTokenizer == null || !stringTokenizer.hasMoreElements()) {
                try {
                    stringTokenizer = new StringTokenizer(bufferedReader.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return stringTokenizer.nextToken();
        }

        String nextLine() {
            String spl = "";
            try {
                spl = bufferedReader.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return spl;
        }
    }
    /*****************************************************************************/
}