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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;


public class A {

    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String[] line = in.readLine().split(" ");
        int n = Integer.parseInt(line[0]);
        int k = Integer.parseInt(line[1]);
        if (k > n || (k == 1 && n > 1)) {
            System.out.println(-1);
            return;
        }
        if (n == 1) {
            System.out.println("a");
            return;
        }
        if (n == 2) {
            System.out.println("ab");
            return;
        }
        char[] res = new char[n];
        for (int i=0; i<n-k+2; i++) {
            res[i] = (char)(i%2 + 'a');
        }
        for (int i=2; i<k; i++) {
            res[n-k+i] = (char)(i + 'a');
        }
        System.out.println(new String(res));
    }
    
}
