Codeforces Round 431 (Div. 1) |
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Finished |

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constructive algorithms

*1600

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A. From Y to Y

time limit per test

1 secondmemory limit per test

256 megabytesinput

standard inputoutput

standard outputFrom 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 *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.

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

12

Output

abababab

Input

3

Output

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.

Codeforces (c) Copyright 2010-2024 Mike Mirzayanov

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