codeforces 288A：Polo the Penguin and Strings

Description

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:

The string consists of n lowercase English letters (that is, the string's length equals n), exactly k of these letters are distinct.
No two neighbouring letters of a string coincide; that is, if we represent a string as s = s₁s₂... s_n, then the following inequality holds, s_i ≠ s_i + 1(1 ≤ i < n).
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... x_p is lexicographically less than string y = y₁y₂... y_q, if either p < q and x₁ = y₁, x₂ = y₂, ... , x_p = y_p, or there is such number r (r < p, r < q), that x₁ = y₁, x₂ = y₂, ... , x_r = y_r and x_r + 1 < y_r + 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



正解：贪心
解题报告：
　　直接每次贪心地让ab不断重复，注意特判一些细节，这题的数据好坑

 1 //It is made by jump~
 2 #include <iostream>
 3 #include <cstdlib>
 4 #include <cstring>
 5 #include <cstdio>
 6 #include <cmath>
 7 #include <algorithm>
 8 #include <ctime>
 9 #include <vector>
10 #include <queue>
11 #include <map>
12 #include <set>
13 using namespace std;
14 typedef long long LL;
15 int n,k,cnt;
16 
17 inline int getint()
18 {
19        int w=0,q=0; char c=getchar();
20        while((c<'0' || c>'9') && c!='-') c=getchar(); if(c=='-') q=1,c=getchar(); 
21        while (c>='0' && c<='9') w=w*10+c-'0', c=getchar(); return q ? -w : w;
22 }
23 
24 inline void work(){
25     n=getint(); k=getint();
26     if(k>n || (k==1 && n>1)) { printf("-1"); return ; }    
27     if(k==1 && n==1) { printf("a"); return ; }
28     cnt=2; k-=2;
29     for(int i=1;i<=n;i++) {
30     if(n-i<k) printf("%c",(char)cnt+'a'),cnt++;
31     else {
32         if(i&1) printf("a"); else printf("b");
33     }
34     }
35 }
36 
37 int main()
38 {
39   work();
40   return 0;
41 }