Little penguin Polo adores strings. But most of all he adores strings of length n.

One day he wanted to find a string that meets the following conditions:

  1. The string consists of n lowercase English letters (that is, the string's length equals n), exactly k of these letters are distinct. 
  2. No two neighbouring letters of a string coincide; that is, if we represent a string as s = s1s2... sn, then the following inequality holds, si ≠ si + 1(1 ≤ i < n). 
  3. Among all strings that meet points 1 and 2, the required string is lexicographically smallest. 



Help him find such string or state that such string doesn't exist.

String x = x1x2... xp is lexicographically less than string y = y1y2... yq, if either p < q and x1 = y1, x2 = y2, ... , xp = yp, or there is such number r (r < p, r < q), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 < yr + 1. The characters of the strings are compared by their ASCII codes.

Input

A single line contains two positive integers n and k (1 ≤ n ≤ 106, 1 ≤ k ≤ 26) — the string's length and the number of distinct letters.

Output

In a single line print the required string. If there isn't such string, print "-1" (without the quotes).

Examples

Input

7 4


Output

ababacd


Input

4 7


Output

-1
//package contest;

import java.io.*;
import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

public class ProblemA {
    BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

    BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));

    int[] readInts() throws IOException {
        String[] strings = reader.readLine().split(" ");
        int[] ints = new int[strings.length];
        for(int i = 0; i < ints.length; i++) {
            ints[i] = Integer.parseInt(strings[i]);
        }
        return ints;
    }

    void solve() throws IOException {
        int[] tt = readInts();
        int n = tt[0], k = tt[1];
        if(n < k || (n > 1 && k == 1)) {
            writer.write("-1");
        }
        else {
            char[] a = new char[n];
            if(k <= 2) {
                for(int i = 0; i < n; i++) {
                    if(i % 2 == 0) a[i] = 'a';
                    else a[i] = 'b';
                }
            }
            else {
                for(int i = 0; i < n - k + 2; i++) {
                    if(i % 2 == 0) a[i] = 'a';
                    else a[i] = 'b';
                }
                for(int i = 2; i < k; i++) {
                    a[n - k + i] = (char)('a' + i);
                }
            }
            writer.write(a);
        }
        writer.flush();
    }

    public static void main(String[] args) throws IOException {
        new ProblemA().solve();
    }
}
