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... sn, then the following inequality holds, si ≠ si + 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... 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

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

//It is made by jump~
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <ctime>
#include <vector>
#include <queue>
#include <map>
#include <set>
using namespace std;
typedef long long LL;
int n,k,cnt;

inline int getint()
{
int w=0,q=0; char c=getchar();
while((c<'0' || c>'9') && c!='-') c=getchar(); if(c=='-') q=1,c=getchar();
while (c>='0' && c<='9') w=w*10+c-'0', c=getchar(); return q ? -w : w;
}

inline void work(){
n=getint(); k=getint();
if(k>n || (k==1 && n>1)) { printf("-1"); return ; }
if(k==1 && n==1) { printf("a"); return ; }
cnt=2; k-=2;
for(int i=1;i<=n;i++) {
if(n-i<k) printf("%c",(char)cnt+'a'),cnt++;
else {
if(i&1) printf("a"); else printf("b");
}
}
}

int main()
{
work();
return 0;
}