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
import java.io.*;
import java.util.StringTokenizer;

public class ProblemA {

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

    private BufferedReader in = null;
    private PrintWriter out = null;
    private StringTokenizer tok = new StringTokenizer("");

    public ProblemA() throws Exception {
        init();
        solve();
        in.close();
        out.close();
    }

    private void solve() throws Exception {
        int n = nextInt();
        int k = nextInt();
        if (k == 1) {
            if (n == 1) {
                out.println("a");
            } else {
                out.println(-1);
            }
            return;
        }
        if (n < k) {
            out.println(-1);
            return;
        }
        char[] a = new char[n];
        for (int i = 0; i < n - (k - 2); ++i) {
            a[i] = (char)('a' + (i % 2));
        }
        for (int i = n - (k - 2); i < n; ++i) {
            a[i] = (char)('c' + i - (n - (k - 2)));
        }
        out.println(a);
    }


    private void init() throws FileNotFoundException {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
    }


    private int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    private long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }


    private String nextToken() throws IOException {
        while (!tok.hasMoreTokens()) {
            String s = in.readLine();
            if (s == null) {
                return null;
            }
            tok = new StringTokenizer(s);
        }
        return tok.nextToken();
    }

}
            