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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class CF288A {
public static void main(String []args) {
      MyScanner in = new MyScanner();
      out = new PrintWriter(new BufferedOutputStream(System.out));
      //Start your solution from here.
      int n = in.nextInt();
      int k = in.nextInt();
      boolean flag = false;
      if(n < k || (n>1 && k==1)){
    	  flag = true;
      }
      boolean f = false;
      if((n==1 && k==1)){
    	  out.print('a');
          f = true;
      }
      if(!flag && !f){
    	  char [] ar = new char[n];
    	  char [] op = new char[26];
    	  op[0] = 'a';
    	  for(int i=1;i<op.length;++i){
    		  op[i] = (char) (op[i-1] + 1);
    	  }
    	  char [] nw = new char[k];
    	  for(int i=0;i<nw.length;++i){
    		  nw[i] = op[i];
    	  }
    	  int l = k - 2;
    	  int m = n - l;
    //	  l--;
    	  boolean fl = false;
    	  for(int i=0;i<m;++i){
    		  if(!fl){
    			  ar[i] = 'a';
    			  fl = true;
    		  }
    		  else{
    			  ar[i] = 'b';
    			  fl = false;
    		  }
    	  }
    	  l = 2;
    	  for(int i=m;i<ar.length;++i){
    		  ar[i] = op[l];
    		  l++;
    	  }
    	  out.println(String.valueOf(ar));
      }
      else if(!f){
    	  out.println(-1);
      }
      
      //End your solution here.       
      out.close();
   }
 
 

   //-----------PrintWriter for faster output---------------------------------
   public static PrintWriter out;

   //-----------MyScanner class for faster input----------
   public static class MyScanner {
      BufferedReader br;
      StringTokenizer st;

      public MyScanner() {
         br = new BufferedReader(new InputStreamReader(System.in));
      }

      String next() {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }

      int nextInt() {
          return Integer.parseInt(next());
      }

      long nextLong() {
          return Long.parseLong(next());
      }

      double nextDouble() {
          return Double.parseDouble(next());
      }
      char nextChar()  {
    	  return next().charAt(0);
      }
      int [] nextIntAr(int [] ar){
    	  for(int i=0;i<ar.length;++i){
    		  ar[i] = Integer.parseInt(next());
    	  }
    	  return ar;
      }
      long [] nextLongAr(long [] ar){
    	  for(int i=0;i<ar.length;++i){
    		  ar[i] = Long.parseLong(next());
    	  }
    	  return ar;
      }
      double [] nextDoubleAr(double [] ar){
    	  for(int i=0;i<ar.length;++i){
    		  ar[i] = Double.parseDouble(next());
    	  }
    	  return ar;
      }
      String [] nextStringAr(String [] ar){
    	  for(int i=0;i<ar.length;++i){
    		  ar[i] = next();
    	  }
    	  return ar;
      }

      String nextLine(){
          String str = "";
      try {
         str = br.readLine();
      } catch (IOException e) {
         e.printStackTrace();
      }
      return str;
      }

   }
}