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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 $100 000$ letters, such that the minimum accumulative cost of the whole process is exactly $*k*$. It can be shown that a solution always exists.

Input

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

Output

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.

Note

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.