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.Arrays;
import java.util.Random;
import java.util.StringTokenizer;

public class codeforces715_D {
    static char[] alphabet = "abcdefghijklmnopqrstuvwxyz".toCharArray();

    private static void solve(FastIOAdapter ioAdapter) {
        int n = ioAdapter.nextInt();
        int k = ioAdapter.nextInt();

        if (k != 1) {
            boolean tobase = true;
            int base = 0;
            int second = 1;
            boolean welcome = true;
            for (int i = 0; i < n; i++) {
                if (welcome) {
                    ioAdapter.out.print(alphabet[base]);
                    if (base == k - 1) {
                        base = 0;
                        second = base + 1;
                        welcome = true;
                    } else {
                        welcome = false;
                    }
                } else {
                    if (tobase) {
                        ioAdapter.out.print(alphabet[base]);
                    } else {
                        ioAdapter.out.print(alphabet[second]);
                        if (second == k - 1) {
                            base++;
                            second = base + 1;
                            welcome = true;
                        } else {
                            second++;
                        }
                    }
                    tobase = !tobase;
                }
            }
            ioAdapter.out.println();
        } else {
            ioAdapter.out.println("a".repeat(n));
        }
    }
    
    public static void main(String[] args) throws Exception {
        try (FastIOAdapter ioAdapter = new FastIOAdapter()) {
            int count = 1;
//            count = ioAdapter.nextInt();
            while (count-- > 0) {
                solve(ioAdapter);
            }
        }
    }

    static void shuffleSort(int[] arr) {
        int n = arr.length;
        Random rnd = new Random();
        for (int i = 0; i < n; ++i) {
            int tmp = arr[i];
            int randomPos = i + rnd.nextInt(n - i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
    }

    static class FastIOAdapter implements AutoCloseable {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        public PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter((System.out))));
        StringTokenizer st = new StringTokenizer("");

        String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

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

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

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

        @Override
        public void close() throws Exception {
            out.flush();
            out.close();
            br.close();
        }
    }
}