题干：

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

题目大意：（摘抄）

给定一个长度为n的字符串，让你用前k种小写字母将字符串排成一个长度为n的，左右相邻字母不相同的，且字典序最小的字符串。注意必须要用K种。如果不能做到则输出-1.

解题报告：

   水题，，但是还是有些粗心的，，k==1的时候也需要特判啊！！

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 1e6 + 5;
int n,k;
char s[MAX];
int main()
{
	cin>>n>>k;
	if(k>n) {
		puts("-1");return 0 ;
	}
	if(n == 1) {
		printf("a");return 0 ;
	}
	if(n>1 && k==1) {
		puts("-1");return 0 ;
	}
	int tmp = k-2;
	int tt=tmp+2;
	for(int i = n; i>=n-tmp+1; i--) {
		s[i] = 'a'-1 + tt;
		tt--;
	}
	for(int i = 1; i<=n-tmp; i++) {
		if(i%2==1) s[i] = 'a';
		else s[i]='b';
	}
	printf("%s\n",s+1);
	return 0 ;
 }