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 C {
    public static Scanner scan = new Scanner(System.in);

    public static void main(String[] args) throws Exception {
        int n = Integer.parseInt(scan.next());
        int k = Integer.parseInt(scan.next());
        if (n<k){
            System.out.println(-1);;
        }
        
        else if (k==1){
            if (n==1){
                System.out.println("a");
            }
            else {
                System.out.println(-1);
            }
        }
        
        else if (k == 2){
            StringBuilder fin = new StringBuilder();
            int c = 1;
            for (int i=0;i<n;i++){
                if (c==1){
                    fin.append('a');
                }
                else {
                    fin.append('b');
                }
                c*=-1;
            }
            System.out.println(fin);
        }
        else {
            StringBuilder fin = new StringBuilder();
            fin.append("ab");
            int c = 1;
            for (int i=0;i<n-k;i++){
                if (c==1){
                    fin.append('a');
                }
                else {
                    fin.append('b');
                }
                c*=-1;
            }
            int ptr = 0;
            for (int i=0;i<k-2;i++){
                fin.append((char)('c'+ptr));
                ptr++;
            }
            System.out.println(fin);
            
        }
    }
}
