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 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.

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

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.

12

abababab

3

codeforces

 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.