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.util.Collections;

public class B {

 /**
  * @param args
  */
 public static void main(String[] args) throws IOException{
  // TODO Auto-generated method stub
  Scanner sc = new Scanner(System.in); 
  int n = sc.nextInt();
  int k = sc.nextInt();
  char[] res = new char[n];
  if( k > n || (k==1 && n>1)) {
      System.out.println("-1");
  } else if(k== 1 && n==1) {
       System.out.println("a");
  }
  else {
      char c = 'c';
      for(int i=n-(k-2); i<n;i++, c++) res[i] = c;
      for(int i=0;i<n-k+2;i++) {
        if(i%2 == 0) res[i] = 'a';
        else res[i] = 'b';
      }
      System.out.println(new String(res));
  }
 }
}
  