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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.

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.

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.

思路：

　　观察样例1很容易发现对于同一种字母，最小花费显然是是n*(n-1)/2，所以可以尝试用几个n*(n-1)/2去构造k。

　　具体怎么证明没多想，用代码测了下发现都是对的，就写了一发。

 #include <bits/stdc++.h>

 using namespace std;

 #define MP make_pair

 #define PB push_back

 typedef long long LL;

 typedef pair<int,int> PII;

 const double eps=1e-;

 const double pi=acos(-1.0);

 const int K=1e6+;

 const int mod=1e9+;

 int k,cnt,ls,num,sum;

 int main(void)

 {

     cin>>k;

     if(k==)

     {

         printf("a\n");

         return ;

     }

     while(k>)

     {

         ls=num=sum=;

         while(k>=sum)

             ls=sum,num++,sum+=num;

         k-=ls;

         for(int i=; i<=num; i++)

             printf("%c",'a'+cnt);

         if(k<=)break;

         cnt++;

     }

     return ;

 }