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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 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
  1. 12
output
  1. abababab
input
  1. 3
output
  1. 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.

题意:可能说的不清楚,我们取两个字符串,重复的我们把出现次数记录一下,然后相乘

a和b没有重复的,0*0

aba和b有一个重复的 1*1

然后。。为什么最后等于8了我也没想(为什么不是4*4或者1*1?)

反正最后我们加起来等于n就行

解法:

1 构造当然想最容易的 n=12

a a a a这种合并就很简单,0+1+2+3就行

2 我们拿5个a,花费了10,还差2

3 换个字母b,拿两个b b ,还差1

4 再换个字母c,两个c c 搞定

  1. #include<bits/stdc++.h>
  2. using namespace std;
  3. double x[];
  4. set<double>Se;
  5. double ans;
  6. int main(){
  7. int n;
  8. cin>>n;
  9. string s="";
  10. if(n==){
  11. cout<<"a"<<endl;
  12. }else{
  13. char c='a';
  14. while(n){
  15. int sum=;
  16. int i=;
  17. for(i=;sum<=n;i++){
  18. sum+=i;
  19. }
  20.  
  21. n-=(sum-i+);
  22. for(int j=;j<i-;j++){
  23. s+=c;
  24. }
  25. c++;
  26. }
  27. cout<<s<<endl;
  28. }
  29. return ;
  30. }

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