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 cf289c {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n=in.nextInt(),k=in.nextInt();
        if(k<2||k>n||k>26){
            if((k==1)&&(n==1))
              System.out.println("a");
            else
            System.out.println("-1");
            return;
        }
        int t=k-2;
        String tail="";
        for(int i=0;i<t;i++){
            char c=Character.toChars('c'+i)[0];
            tail=tail+c;
        }
        int d=n-t;
        if(d%2==1)tail=""+'a'+tail;
        d=d/2;
        StringBuilder sb=new StringBuilder(d);
        while(d>0){sb.append("ab");d--;}
        sb.append(tail);
        System.out.println(sb);
    }
}

