A11086. From Y to Y

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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 $100000$ letters, such that the minimum accumulative cost of the whole process is exactly $k$ . It can be shown that a solution always exists.

The first and only line of input contains a non-negative integer $k$ ( $0<=k<=100000$ ) — the required minimum cost.

Output a non-empty string of no more than $100000$ 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.

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.