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
//package ru.akvasov.codeforces.r244;

import java.util.Scanner;

/**
 * Created by akvasov on 15.05.14.
 */
public class A {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        int n = scanner.nextInt();
        int k = scanner.nextInt();

        StringBuilder sb = new StringBuilder();

        if (n == 1 && k == 1){
            System.out.println("a");
            return;
        }

        if (k < 2 || k > n) {
            System.out.println(-1);
            return;
        }

        int t = n - k + 2;

        for (int i = 0; i < t; i++){
            if (i % 2 == 0) {
                sb.append("a");
            } else {
                sb.append("b");
            }
        }

        if (t >=0){
            for (int i = 0; i < k - 2; i++){
                sb.append((char)('c' + i));
            }
        }

        String ans = sb.toString();
        if (ans.isEmpty()){
            System.out.println(-1);
        } else {
            System.out.println(sb.toString());
        }

    }
}
