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.util.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStream;

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

class TaskA {

    private String getCharForNumber(int i) {
        return i > 0 && i < 27 ? String.valueOf((char)(i + 'a' - 1)) : null;
    }

    public void solve(int testNumber, Scanner in, PrintWriter out) {
        int n = in.nextInt();
        int k = in.nextInt();
        if (k > n || (k == 1 && n!=1)) {
            out.println(-1);
            return;
        }
        if (n == 1) {
            out.print("a");
            return;
        }
        int count = 0;
        int i = 0;
        while (count < n-(k)+2) {
            out.print(getCharForNumber(i+1));
            i = (i + 1) % 2;
            count++;
        }
        i = 2;
        while (count < n) {
            out.print(getCharForNumber(i+1));
            i = (i + 1) % k;
            count++;
        }
    }
}

