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.*;
import java.io.*;

public class Main implements Runnable {
		
	public void solve() throws IOException {
		int L = nextInt();
                int letters = nextInt();
                if(L == 1 && letters == 1) {
                        System.out.println('a');
                        return;
                }
                
                if(letters == 1 && L > 1){
                        System.out.println(-1);
                        return;
                }
                StringBuilder sb = new StringBuilder();
                if(letters > L ){
                        System.out.println(-1);
                        return;
                }
                
                int now = L - (letters - 2);
                boolean a = true;
                for(int i = 0; i < now; i++){
                        if(a) sb.append('a');
                        else sb.append('b');
                        
                        a = !a;
                }
                
                for(int i = 2; i < letters; i++) sb.append((char)('a' + i));
                
                System.out.println(sb.toString());
	}

	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	
	//-----------------------------------------------------------
	public static void main(String[] args) {
		new Main().run();
	}

	public void run() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			tok = null;
			solve();
			in.close();
		} catch (IOException e) {
			System.exit(0);
		}
	}

	public String nextToken() throws IOException {
		while (tok == null || !tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine());
		}
		return tok.nextToken();
	}

	public int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	public long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	public double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	BufferedReader in;
	StringTokenizer tok;
}