題目：

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 x

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

解題報告：輸入的n，k，n是代表字串的長度，k是字串中出現的字母的次序（從a，b，c……），問怎麼尋找最小字典序的字串，根據題目樣例，能夠找到幾個坑點，k=1是若n！=1 ，無解，k=0||k>26無解。

剩餘的情況就是 將n-k+1的前邊用ab組成填滿，然後後邊的用cdef……升序組成。

ac程式碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;

const int maxn =1e6+100;

char str[maxn];

int main()
{
	int n,k;
	while(cin>>n>>k)
	{
		if(n<k||k<=0||k>26)
		{
			printf("-1\n");
			continue;
		}
		if(n==1&&k==1)
		{
			printf("a\n");
			continue;
		}
		else if(k==1&&n!=1)
		{
			printf("-1\n");
			continue;
		}
		else
		{
			str[0]='a',str[1]='b';
			int i;
			for(i=2;i<=n-k+1;i++)
			{
				if(i%2==0)
					str[i]='a';
				else
					str[i]='b';
			}
			for(int j=2;j<k;j++,i++)
			{
				str[i]='a'+j;
			}
			str[n]='\0';
			cout<<str<<endl;
		}
			
	}
}