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.*;

public class C {
  public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int n = in.nextInt();
    int k = in.nextInt();
    if (n < k || (n >= 2 && k < 2)) {
      System.out.println(-1);
    } else if (n == 1 && k == 1) {
      System.out.println("a");
    } else if ((n - k ) % 2 == 1) {
      StringBuilder sb = new StringBuilder();
      for (int i = 0; i < n - k + 1; i++) {
        sb = sb.append(i % 2 == 0 ? "a" : "b");
      }
      sb.append("a");
      for (int i = 0; i < k - 2; i++) {
        sb = sb.append((char)('c' + i));
      }
      System.out.println(sb.toString());
    } else {
      StringBuilder sb = new StringBuilder();
      for (int i = 0; i < n - k + 2; i++) {
        sb = sb.append(i % 2 == 0 ? "a" : "b");
      }
      for (int i = 0; i < k - 2; i++) {
        sb = sb.append((char)('c' + i));
      }
      System.out.println(sb.toString());
    }
    in.close();
  }
}
