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;

public class pengString {

    public static void main(String args[]) {
        Scanner stdin = new Scanner(System.in);
        int n = stdin.nextInt();
        int k = stdin.nextInt();
        if (n < k || (k==1 && n>1)) {
            System.out.print(-1);
        } else if (k == 1) {
            System.out.print("a");
        } else {
            char[] out=new char[k-2];
            for (int j = 99; j < (99 + k - 2); j++) {
                out[j-99] = (char) j;
            
            }
            
            int abPair = n - (k - 2);
            char[] abString = new char[abPair];
            for (int q = 0; q < abPair; q++) {
                abString[q]=(char) (97+(q%2));
            }
            String first=new String(abString);
            String second=new String(out);
            System.out.print(first+second);
        }

    }
}