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.*;

public class A{

    static String solve(int n, int k) {
        if (n < k) return "-1";
        StringBuilder sb = new StringBuilder();
        if (n == k)
            for (int i=0; i<k; i++)
                sb.append(((char)( ((int)'a') + i)));
        else if (k == 1) // How can I be so stupid!
            return "-1";
        ///for (int i=0; i<n; i++)
        // sb.append('a');
        else { // n >= 3, k >= 2
            for (int i=0; i<n-k+2; i++)
                if (i%2 == 0) sb.append('a');
                else sb.append('b');
            for (int j=0; j<k-2; j++)
                sb.append( ((char) ((int)'c' + j)));
        }
        return sb.toString();
    }

    public static void main(String [] argv) {
        Kattio io = new Kattio(System.in, System.out);
        int n = io.getInt();
        int k = io.getInt();
        io.println(solve(n, k));
        io.close();
    }

}


// ####### KATTIO ########
class Kattio extends PrintWriter {
    public Kattio(InputStream i) {
        super(new BufferedOutputStream(System.out));
        r = new BufferedReader(new InputStreamReader(i));
    }
    public Kattio(InputStream i, OutputStream o) {
        super(new BufferedOutputStream(o));
        r = new BufferedReader(new InputStreamReader(i));
    }

    public boolean hasMoreTokens() {
        return peekToken() != null;
    }

    public int getInt() {
        return Integer.parseInt(nextToken());
    }

    public double getDouble() {
        return Double.parseDouble(nextToken());
    }

    public long getLong() {
        return Long.parseLong(nextToken());
    }

    public String getWord() {
        return nextToken();
    }



    private BufferedReader r;
    private String line;
    private StringTokenizer st;
    private String token;

    private String peekToken() {
        if (token == null)
            try {
                while (st == null || !st.hasMoreTokens()) {
                    line = r.readLine();
                    if (line == null) return null;
                    st = new StringTokenizer(line);
                }
                token = st.nextToken();
            } catch (IOException e) { }
        return token;
    }

    private String nextToken() {
        String ans = peekToken();
        token = null;
        return ans;
    }
}
