A. Polo the Penguin and Strings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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).

Sample test(s)

input

7 4

output

ababacd

input

4 7

output

-1

解题说明：此题是一道典型的贪心问题。既要保证字符串中随意两个连续字符串不同。也要保证字典序最小。最简单的想法是仅仅用a,b交替，在最后补上其它字符串就可以。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include <algorithm>
#include<cstring>
#include<string>
using namespace std;

int main()
{
	int n, k, c, r, i;
	scanf("%d%d", &n, &k);
	if (k == n&&k <= 26)
	{
		c = 'a';
		for (i = 0; i<k; i++)
		{
			printf("%c", c + i);
		}
		printf("\n");
	}
	else if (k>n || k == 1)
	{
		printf("-1\n");
	}
	else
	{
		for (i = 0; i<n - (k - 2); i++)
		{
			if (i % 2 == 0)
			{
				printf("a");
			}
			else
			{
				printf("b");
			}
		}
		r = 2; 
		c = 'a';
		for (i; i < n; i++, r++)
		{
			printf("%c", c + r);
		}
		printf("\n");
	}
	return 0;
}