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.InputStreamReader;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author ocelopilli
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskC {
    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int n = in.nextInt();
        int k = in.nextInt();
        if ( n < k ) out.println( -1 );
        else
        {
            if ( k == 1 )
            {
                if ( n == 1 ) out.println("a");
                else out.println(-1);
            }
            else
            {
                char[] ans = new char[ n ];

                for (int i=0; i<n; i++) ans[i] = (i % 2 == 0) ? 'a' : 'b';
                while ( k > 2 ) ans[ --n ] = (char)(--k + 'a');

                out.println( new String( ans ) );
            }
        }
    }
}

class InputReader {
    BufferedReader br;
    StringTokenizer st;

    public InputReader(InputStream in)
    {
        br = new BufferedReader(new InputStreamReader(in));
        st = null;
    }

    public String next()
    {
        while (st==null || !st.hasMoreTokens())
        {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return st.nextToken();
    }

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