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: Accepted)

#![allow(unused_imports)]

use std::cmp::{max, min, Ordering};
use std::collections::*;
use std::{i32, i64, i128, u32, u64, u128, f64};
use std::io::{stdout, Write, BufWriter};
use std::mem::{swap};
use proconio::marker::{Isize1, Usize1, Bytes, Chars};

#[proconio::fastout]
fn main() {
    proconio::input!{
        mut n: i64,
    }
    let mut ans = String::new();
    let mut k = 0;
    for i in 0.. {
        if 26i64.pow(i) <= 25*n + 1 && 25*n + 1 < 26i64.pow(i + 1) {
            k = i;
            n -= (26i64.pow(k) - 1) / 25;
            break;
        }
    }
    for i in (0..k).rev() {
        let q = (n / 26i64.pow(i)) as u8;
        ans.push(char::from(q + b'a'));
        n %= 26i64.pow(i);
    }
    println!("{}", ans);
}