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.InputStreamReader;
import java.util.StringTokenizer;


public class round177C {

    static BufferedReader br = new BufferedReader(new InputStreamReader(
            System.in));
    static StringTokenizer st = new StringTokenizer("");

    static int nextInt() throws Exception {
        return Integer.parseInt(next());
    }

    static String next() throws Exception {
        while (true) {
            if (st.hasMoreTokens()) {
                return st.nextToken();
            }
            String s = br.readLine();
            if (s == null) {
                return null;
            }
            st = new StringTokenizer(s);
        }
    }
    
    public static void main(String[] args)throws Exception {
        int n = nextInt();
        int k = nextInt();
        if(k > n){
            System.out.println(-1);
        }else{
            if(n == 1 && k == 1){
                System.out.println("a");
                return;
            }
            if(n == 2){
                if(k == 1){
                    System.out.println(-1);
                }
                if(k == 2){
                    System.out.println("ab");
                }
                return;
            }           
            char [] ans = new char [n];
            if(k == 1){             
                System.out.println(-1);
                return;
            }
            int rem = k - 2;
            int i;
            for(i = 0 ; i < n - rem ; ++i)
                if(i % 2 == 0)
                    ans[i] = 'a';
                else
                    ans[i] = 'b';
            int off = 2;
            while(i < n)
                ans[i++] = (char) ('a' + off++);
            System.out.println(new String(ans));
        }
    }
}
