CUIT ACM Personal Training 11.27(FM）F - Polo the Penguin and Strings

F - Polo the Penguin and Strings

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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:

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... s_n, then the following inequality holds, s_i ≠ s_i + 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... x_p is lexicographically less than string y = y₁y₂... y_q, if either p < q and x₁ = y₁, x₂ = y₂, ... , x_p = y_p, or there is such number r (r < p, r < q), that x₁ = y₁, x₂ = y₂, ... , x_r = y_r and x_r + 1 < y_r + 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

題解：這個題的題目長度不短，具體講的是給你k個不同的字母，要你輸出n長度的字符串，要求：1.字符串字典序最小；2.字符串相鄰的字母不能相同，這個就是一個貪心算法。首先爲了字典序最小，所以k個字母肯定是從前開始的，選定了字母之後，又要相鄰的字符串不相同，那麼當k=1，也就是隻有1個字母的時候，除非n=1，只輸出一個a，否則就只能輸出-1。k!=1,時，當n<k,則表示用k個字母去組成一個比字母的個數還要少的字符串，這明顯是不成立的，所以輸出-1。n>k時，那麼就一直輸出ab，直到最後剩下k-2個位子放餘下的字母便行了。

AC代碼如下：

#include<bits/stdc++.h>
using namespace std;

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