CF848A.From Y to Y

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题目描述

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

For a given unordered multiset of nn lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 11 , and repeat the following operation n1n-1 times:

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

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

Given a non-negative integer kk , construct any valid non-empty set of no more than 100000100000 letters, such that the minimum accumulative cost of the whole process is exactly kk . It can be shown that a solution always exists.

输入格式

The first and only line of input contains a non-negative integer kk ( 0<=k<=1000000<=k<=100000 ) — the required minimum cost.

输出格式

Output a non-empty string of no more than 100000100000 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.

输入输出样例

  • 输入#1

    12
    

    输出#1

    abababab
    
  • 输入#2

    3
    

    输出#2

    codeforces
    

说明/提示

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 00 ;
  • {"aba", "b", "a", "b", "a", "b"}, with a cost of 11 ;
  • {"abab", "a", "b", "a", "b"}, with a cost of 11 ;
  • {"abab", "ab", "a", "b"}, with a cost of 00 ;
  • {"abab", "aba", "b"}, with a cost of 11 ;
  • {"abab", "abab"}, with a cost of 11 ;
  • {"abababab"}, with a cost of 88 .

The total cost is 1212 , and it can be proved to be the minimum cost of the process.

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