Problem:
Score : 300 points

Problem Statement

1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:

the dogs numbered 1,2,\cdots,26 were respectively given the names a, b, ..., z;

the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names aa, ab, ac, ..., zy, zz;

the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names aaa, aab, aac, ..., zzy, zzz;

the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names aaaa, aaab, aaac, ..., zzzy, zzzz;

the dogs numbered 475255,475256,\cdots were respectively given the names aaaaa, aaaab, ...;

and so on.

To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:

a, b, ..., z, aa, ab, ..., az, ba, bb, ..., bz, ..., za, zb, ..., zz, aaa, aab, ..., aaz, aba, abb, ..., abz, ..., zzz, aaaa, ...

Now, Roger asks you:

"What is the name for the dog numbered N?"

Constraints

N is an integer.

1 \leq N \leq 1000000000000001

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer to Roger's question as a string consisting of lowercase English letters.

Sample Input 1

2

Sample Output 1

b

Sample Input 2

27

Sample Output 2

aa

Sample Input 3

123456789

Sample Output 3

jjddja

Solution in Rust (1.42.0) (judge verdict: Wrong Answer)

#![allow(unused_imports)]

//input
use proconio::input;
use proconio::marker::{Bytes, Chars};
//use std::io::Read;

//collections
//use std::collections::BTreeMap;
//use std::collections::BTreeSet;
//use std::collections::HashMap;
//use std::collections::HashSet;
use std::collections::VecDeque;
//use std::collections::BinaryHeap;

//use std::cmp::*;

fn main() {
    input! {
        mut n: u64,
    }

    let mut ans = VecDeque::new();
    for i in 1.. {
        if n == 0 {
            break;
        }
        let tmp = n % 26u64.pow(i);
        if tmp != 0 {
            ans.push_front(tmp);
            n -= tmp;
        }
    }

    for i in 0..ans.len() {
        //println!("{}", (ans[i] / 26u64.pow(i as u32)));
        let v = ans[i] / 26u64.pow((ans.len() - i - 1) as u32);
        print!("{}", (v as u8 + 'a' as u8 - 1) as char);
    }
    println!();
}