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.*;

public class Main288A
{
  static PrintWriter out=new PrintWriter(System.out);
  
  public static void main(String[] args) throws IOException
  {
    Scanner sc=new Scanner(System.in);
    int n=sc.nextInt();
    int k=sc.nextInt();
    if(k>n||(k<=1&&n>1))
      out.println(-1);
    else if(k==1&&n==1)
      out.println("a");
    else
    {
      char[] ans=new char[n];
      int p=0;
      int x=n-(k-2);
      for(int i=0;i<x;i++)
      {
        if(i%2==0)
          ans[p++]='a';
        else
          ans[p++]='b';
      }
      int c=(int)'b';
      for(int i=0;i<(k-2);i++)
      {
        ans[p++]=(char)(c+1+i);
      }
      out.println(ans);
    }
    out.flush();
        
    
    
  }
  static class Scanner
  {
    BufferedReader br;
    StringTokenizer tk=new StringTokenizer("");
    public Scanner(InputStream is) 
        {
      br=new BufferedReader(new InputStreamReader(is));
    }
        public int nextInt() throws IOException
        {
          if(tk.hasMoreTokens())
                return Integer.parseInt(tk.nextToken());
          tk=new StringTokenizer(br.readLine());
          return nextInt();
        }
        public long nextLong() throws IOException
        {
            if(tk.hasMoreTokens())
              return Long.parseLong(tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return nextLong();
        }
        public String next() throws IOException
        {
          if(tk.hasMoreTokens())
            return (tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return next();
        }
        public String nextLine() throws IOException
        {
          tk=new StringTokenizer("");
          return br.readLine();
        }
        public double nextDouble() throws IOException
        {
            if(tk.hasMoreTokens())
              return Double.parseDouble(tk.nextToken());
            tk=new StringTokenizer(br.readLine());
            return nextDouble();
        }
        public char nextChar() throws IOException
        {
            if(tk.hasMoreTokens())
              return (tk.nextToken().charAt(0));
            tk=new StringTokenizer(br.readLine());
            return nextChar();
        }
        public int[] nextIntArray(int n) throws IOException
        {
            int a[]=new int[n];
            for(int i=0;i<n;i++)
              a[i]=nextInt();
            return a;
        }
        public long[] nextLongArray(int n) throws IOException
        {
          long a[]=new long[n];
            for(int i=0;i<n;i++)
              a[i]=nextLong();
            return a;
        }
        public int[] nextIntArrayOneBased(int n) throws IOException
        {
          int a[]=new int[n+1];
            for(int i=1;i<=n;i++)
              a[i]=nextInt();
            return a;
        }
        public long[] nextLongArrayOneBased(int n) throws IOException
        {
          long a[]=new long[n+1];
          for(int i=1;i<=n;i++)
                a[i]=nextLong();
          return a;
        }
  }
}

