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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class C {
    
    static StringTokenizer st;
    static BufferedReader in;
    public static void main(String[] args) throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int n = nextInt();
        int k = nextInt();
        if (k > n) {
            System.out.println(-1);
            return;
        }
        if (k==1) {
            if (n==1)
                System.out.println("a");
            else
                System.out.println(-1);
            return;
        }
        boolean[]used = new boolean[27];
        int noteused = k;
        for (int i = 1; i <= n; i++) {
            if (n-i+1 <= noteused) {
                for (int j = 1; j <= k; j++) {
                    if (!used[j])
                        pw.print((char)(j+96));
                }
                break;
            }
            else {
                if (i % 2==1) {
                    pw.print('a');
                    used[1] = true;
                    if (noteused==k)
                        noteused = k-1;
                }
                else {
                    pw.print('b');
                    used[2] = true;
                    noteused = k-2;
                }
            }
        }
        pw.close();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }
    
    private static double nextDouble() throws IOException{
        return Double.parseDouble(next());
    }
    
    private static String next() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
}
