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: Accepted)

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef long int lt;
typedef vector<ll> vi;
typedef vector<int> vin;
typedef pair<ll,ll> pll;
typedef pair<int,int> pin;
#define all(a) a.begin(),a.end()
#define rep(i,a,b) for(int i=a;i<b;++i)
#define f(i,b) for(int i=0;i<b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define test(t) int t; cin>>t; while(t--) solve()
#define endl "\n"
#define p1 cout<<"YES"<<endl
#define p0 cout<<"NO"<<endl
#define fst ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)

const int mod=1e9+7;

ll mod_pow(ll a,ll n,ll mod)	//(a^n)%mod
{	ll res=1;
	while(n){	if(n&1)	res=(res*a)%mod; n/=2; a=(a*a)%mod;	}
	return res;
}

void solve();

int main()
{
    fst;
    //test(t);
     solve();
}

/**********************<="Actual code strats here"=>*****************************/

ll k,n;
char getch()
{
    int a=k%26;
    if(a==0) {k-=26;return 'z';}
    else {k-=a; return (char)(a-1+'a');}
}

void solve()
{
    cin>>k;

    string ans="";

    while(k>0)
    {
        ans+=getch();
        k/=26;
    }

    reverse(all(ans));
    cout<<ans<<endl;
}