CF848A.From Y to Y

From beginning till end, this message has been waiting to be conveyed.

For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:

• Remove any two elements s and t from the set, and add their concatenation s + t to the set.

The cost of such operation is defined to be , where f(s, c) denotes the number of times character c appears in string s.

Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.

从开始到结束，这条消息一直在等待被传达。

给定一个由 $ n $ 个小写英文字母组成的无序多重集（“多重”表示某个字母可能出现多次），我们将所有字母视作长度为 1 的字符串，并重复执行以下操作 $ n-1 $ 次：

• 从集合中任选两个元素 $ s $ 和 $ t $ 并将其移除，然后将它们的连接串 $ s + t $ 加入集合。

该操作的代价定义为 ，其中 $ f(s,,c) $ 表示字符 $ c $ 在字符串 $ s $ 中出现的次数。

给定一个非负整数 $ k $，请构造任意一个合法的、非空且包含至多 $ 100,000 $ 个字母的集合，使得整个过程的最小累计代价恰好为 $ k $。可以证明，这样的解总是存在的。

The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.

输入仅有一行，包含一个非负整数 $k$（$0 \leq k \leq 100\,000$）——即所要求的最小代价。

Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.

Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.

输出一个非空字符串，长度不超过 100 000，且仅由小写英文字母组成——即任意一个满足要求的多重集，将其元素连接而成的字符串。

注意：所输出的字符串无需是最终拼接得到的字符串；它只需表示一个字母的无序多重集。

For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:

• {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
• {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
• {"abab", "a", "b", "a", "b"}, with a cost of 1;
• {"abab", "ab", "a", "b"}, with a cost of 0;
• {"abab", "aba", "b"}, with a cost of 1;
• {"abab", "abab"}, with a cost of 1;
• {"abababab"}, with a cost of 8.

The total cost is 12, and it can be proved to be the minimum cost of the process.

对于多重集 $\{'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'\}$，完成该过程的一种方式如下：

• $\{"ab", "a", "b", "a", "b", "a", "b"\}$，代价为 $0$；
• $\{"aba", "b", "a", "b", "a", "b"\}$，代价为 $1$；
• $\{"abab", "a", "b", "a", "b"\}$，代价为 $1$；
• $\{"abab", "ab", "a", "b"\}$，代价为 $0$；
• $\{"abab", "aba", "b"\}$，代价为 $1$；
• $\{"abab", "abab"\}$，代价为 $1$；
• $\{"abababab"\}$，代价为 $8$。

总代价为 $12$，且可以证明这是该过程的最小代价。