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;
 
#define ll long long 
#define ld long double
#define pii pair<int,int>
#define pll pair<ll,ll>
#define mkp make_pair
#define pb push_back
#define pf push_front
#define popb pop_back
#define popf pop_front
#define ff first
#define ss second
#define endl "\n"
#define MOD 1000000007
#define MOD1 998244353
#define inf 1e18
 
ll GCD(ll x,ll y){if(y==0)return x;return GCD(y,x%y);}
ll LCM(ll x,ll y){return (x*y)/(GCD(x,y));}
ll LOGK(ll x,ll k){if(x>=k)return 1+LOGK(x/k,k);return 0;}
ll MPOW( ll a, ll b, ll m) { if(b==0) return 1; ll x=MPOW(a,b/2,m); x=(x*x)%m; if(b%2==1) x=(x*a)%m; return x;}
ll MINV(ll a,ll m) {return MPOW(a,m-2,m);}
 
class pnc{
  ll FACT_MAX,MODU;
  vector<ll> fact;
public:
  pnc( ll n, ll m) { FACT_MAX = n; fact.resize(FACT_MAX); MODU = m; MFACT_INIT(MODU);}
  void MFACT_INIT( ll m) { fact[0]=1; for(ll i=1;i<FACT_MAX;++i) fact[i]=(i*fact[i-1])%MODU;}
  ll MFACT( ll n) { return fact[n];}
  ll PERM( ll n, ll r) { return (fact[n]*::MINV(fact[n-r],MODU))%MODU;}
  ll COMB( ll n , ll r) { return (PERM(n,r)*::MINV(fact[r],MODU))%MODU;}
};
 
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);    
 
    ll test=1;
    // cin>>test;
    while(test--)
    {
        ll n;
        cin>>n;

        string str="abcdefghijklmnopqrstuvwxyz";
        vector<ll> ans;
        while(n!=0)
        {
            n--;
            ans.pb(n%26);
            n/=26;
        }

        reverse(ans.begin(),ans.end());

        for(auto i:ans)
        cout<<str[i];
    }
}