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.io.*;
public class polo {
    public static void main(String[] args)throws IOException {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        String [] line = bf.readLine().split(" ");
        int len = Integer.parseInt(line[0]);
        int letters = Integer.parseInt(line[1]);
        if(letters>len || (letters==1 && len>1)){System.out.println(-1);return;}
        if(len==1 && letters == 1){System.out.println("a");return;}
        for(int i = 0; i<(len-(letters-2))/2;i++){
            System.out.print("ab");
        }
        String c = "";
        switch(letters){
        case 3: c = "c";break;
        case 4: c = "cd";break;
        case 5: c = "cde";break;
        case 6: c = "cdef";break;
        case 7: c = "cdefg";break;
        case 8: c = "cdefgh";break;
        case 9: c = "cdefghi";break;
        case 10: c = "cdefghij";break;
        case 11: c = "cdefghijk";break;
        case 12: c = "cdefghijkl";break;
        case 13: c = "cdefghijklm";break;
        case 14: c = "cdefghijklmo";break;
        case 15: c = "cdefghijklmno";break;
        case 16: c = "cdefghijklmnop";break;
        case 17: c = "cdefghijklmnopq";break;
        case 18: c = "cdefghijklmnopqr";break;
        case 19: c = "cdefghijklmnopqrs";break;
        case 20: c = "cdefghijklmnopqrst";break;
        case 21: c = "cdefghijklmnopqrstu";break;
        case 22: c = "cdefghijklmnopqrstuv";break;
        case 23: c = "cdefghijklmnopqrstuvw";break;
        case 24: c = "cdefghijklmnopqrstuvwx";break;
        case 25: c = "cdefghijklmnopqrstuvwxy";break;
        case 26: c = "cdefghijklmnopqrstuvwxyz";break;
        default: break;
        }
        if((len-letters)%2!=0){System.out.print("a"+c);}
        else{System.out.print(c);}
        
        

    }

}
