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 C {
        
        public static void main(String[] args) 
        {
            Scanner in = new Scanner(System.in);
            int n = in.nextInt();
            int k = in.nextInt();
            if (k > n)
            {
                System.out.println("-1");
                return;
            }
            if (k == 1)
            {
                if (n == 1)
                    System.out.println("a");
                else
                    System.out.println("-1");
                return;
            }
            
            StringBuilder ret = new StringBuilder();
            char cur = 'a';
            for (int i = 0; i < n - (k - 2); i++)
            {
                ret.append(cur);
                cur = cur == 'a' ? 'b' : 'a';
            }
            cur = 'c';
            for (int i = n - (k - 2); i < n; i++)
            {
                ret.append(cur);
                cur++;
            }
            System.out.println(ret.toString());
        }
    }
