SDKD 2016 Summer Single Contest #13 .C

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 Input

Input
7 4

Output
ababacd

Input
4 7

Output
-1
题目：
要求1:字符串中必须出现k个不同的 字母
要求2:不能连续两个相同
要求3:字典序最小
这样的话 前面 全是a最好 但是有要求2
所以 全是ab 就好， 有要求 一 就要 倒着 把剩下的没出现的字母输出，然后剩下的就是ab了

#include <iostream>#include <cstdio>#include <string>#include <cmath>#include <algorithm>using namespace std;int main(){    int n, k;    scanf("%d%d", &n, &k);    //k = 1 时比较特殊     if(n == 1 && k == 1)        printf("a");    else if(n < k || k == 1)        printf("%d\n", -1);    else    {    //k为种类  去掉ab两种 剩下的：        int a = n - (k-2);        for(int i = 1; i <= a; i++)        {//判断奇偶            if(i % 2 == 1)                printf("a");            else                printf("b");        }        int j = 2;        for(int i = a+1; i <= n; i++)        {            printf("%c", 'a'+j);            j++;        }    }    return 0;}