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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)


#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
#define ll long long
#define ld long double
#define MOD 1000000007
#define nl cout << bl
#define rt ll Q[100005] = {0}
#define mp make_pair
#define test() \
    ull t;     \
    cin >> t;  \
    while (t--)
#define pb push_back
#define ff first
#define ss second
#define all(v) v.begin(), v.end()
#define s(ar, n) sort(ar, ar + n)
#define rs(ar, n) sort(ar, ar + n, greater<ll>())
#define oa(a, n)               \
    for (ll i = 0; i < n; i++) \
        cout << a[i] << " ";
#define cn(a, n)               \
    for (ll i = 0; i < n; i++) \
        cin >> a[i];
#define ov(a, n)                    \
    for (ll i = n - 1; i >= 0; i--) \
        cout << a[i] << " ";        \
    nl
#define ovp(a, n)              \
    for (ll i = 0; i < n; i++) \
        cout << a[i].ff << " " << a[i].ss << bl;
#define maxa(ar, N) *max_element(ar, ar + N)
#define mina(ar, N) *min_element(ar, ar + N)
#define present(c, x) ((c).find(x) != (c).end())
#define fastio()                  \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);                   \
    cout.tie(0)
typedef pair<int, int> pii;
ll fast_exp(ll base, ll exp)
{
    ll res = 1;
    while (exp > 0)
    {
        if (exp % 2 == 1)
            res = (res * base) % MOD;
        base = (base * base) % MOD;
        exp /= 2;
    }
    return res % MOD;
}
ll fib(ll n)
{
    if (n < 2)
        return n;
    ll a = 0, b = 1, pre;
    int i = 1;
    while (i < n)
    {
        pre = (a + b) % MOD;
        a = b;
        b = pre;
        i++;
    }
    return pre;
}
ll fac(ll n)
{
    if (n % 2 == 0)
        return 2;

    for (ll i = 3; i * i <= n; i += 2)
    {
        if (n % i == 0)
            return i;
    }

    return n;
}
ll fact(ll n)
{
    if (n > 1)
        return n * fact(n - 1);
    else
        return 1;
}
bool sortbysec(const tuple<int, int, int> &a,
               const tuple<int, int, int> &b)
{
    return (get<1>(a) < get<1>(b));
}
#define MAX 100000
void printString(int n)
{
    char str[MAX]; // To store result (Excel column name)
    int i = 0;     // To store current index in str which is result

    while (n > 0)
    {
        // Find remainder
        int rem = n % 26;

        // If remainder is 0, then a 'Z' must be there in output
        if (rem == 0)
        {
            str[i++] = 'z';
            n = (n / 26) - 1;
        }
        else // If remainder is non-zero
        {
            str[i++] = (rem - 1) + 'a';
            n = n / 26;
        }
    }
    str[i] = '\0';

    // Reverse the string and print result
    reverse(str, str + strlen(str));
    cout << str << endl;

    return;
}
int main()
{
    ll n;
    cin >> n;
    printString(n);
}