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.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;


public class C {

	
	public static void main(String[] args) throws IOException {
InputReader reader=new InputReader();

		int n=reader.nextInt();
		int k=reader.nextInt();
		char c[]={'a','b'};
		
		if(k>n)
			System.out.println(-1);
		else if(n>1 && k==1)
			System.out.println(-1);
		else if(n==1)
			System.out.println('a');
		else{
			StringBuilder b=new StringBuilder();
			b.append(c[0]);
			b.append(c[1]);
			for(int i=0;i<n-k;i++)
			{
				b.append(c[i&1]);
			}
		//	System.out.println(b);
			for(int i=2;i<k;i++)
				b.append((char)('a'+i));
			System.out.println(b.toString());
		}
		
	}
	
	 static class InputReader {
	        BufferedReader in;
	        StringTokenizer st;

	        public InputReader() throws IOException {
	            in = new BufferedReader(new InputStreamReader(System.in));
	            st = new StringTokenizer(in.readLine());

	        }

	        public String next() throws IOException {

	            while (!st.hasMoreElements())
	                st = new StringTokenizer(in.readLine());
	            return st.nextToken();
	        }

	        public int nextInt() throws NumberFormatException, IOException {
	            return Integer.parseInt(next());
	        }

	        public long nextLong() throws NumberFormatException, IOException {
	            return Long.parseLong(next());
	        }
	    }

}
