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
// package Virtual289;

import java.util.Scanner;

public class V289C {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        int k = s.nextInt();
        if(k == n){
            StringBuilder ans = new StringBuilder();
            char ch = 'a';
            for (int i = 0; i < n; i++) {
                ans.append(ch);
                ch++;
            }
            System.out.println(ans.toString());
            return;
        }

        if(k > n || k == 1){
            System.out.println(-1);
            return;
        }



        char ch = 'c';
        StringBuilder ans = new StringBuilder();
        for (int i = 0; i < k - 2; i++) {
            ans.append(ch);
            ch++;
        }
        ans = ans.reverse();
        if((n-k+2) % 2 == 0){
            ch = 'b';
        }else{
            ch = 'a';
        }
        for (int i = k - 2; i < n; i++) {

            ans.append(ch);
            if(ch == 'b'){
                ch = 'a';
            }else{
                ch = 'b';
            }
        }
        System.out.println(ans.reverse().toString());
    }
}
