C. From Y to Y

C. From Y to Y
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 cappears 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.

Examples
input

output

input

output

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.

发现最低费用肯定一个一个加上去,所以直接顺序搞一搞。

 

2017年9月2日 0 / /
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