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.*;
import java.io.*;
import java.math.*;
public class Class{

    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int k = Integer.parseInt(st.nextToken());
        char[] c = new char[n];
        if((k==1&&n>1)||k>n){System.out.print(-1); return;}
        if(k==1&&n==1){System.out.print('a'); return;}
        for(int i=0; i<n-k+2; i++){
            if(i%2==0)c[i]='a';else c[i]='b';
        }
        for(int i=n-k+2; i<n; i++)c[i] = (char)('c'+i-(n-k+2));
        System.out.print(c);
    }
    
    static class Pair{
        long start;
        long end;
        
        public Pair(long start, long end){
            this.start = start;
            this.end = end;
        }
        
        public int hashCode() {
            return (int)(this.start + this.end);
        }
        
        public boolean equals(Object p){
            if(((Pair)p).start==this.start && ((Pair)p).end==this.end)return true;
            return false;
        }
    }
}