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>
#define REP(i, n) for(int i = 0; i < (n); i++)
#define REP1(i, n) for(int i = 1; i <= (n); i++)
#define REPD(i,a,b) for (int i=(a);i<=(b);i++)
#define ALL(v) (v).begin(), (v).end()
using namespace std;
typedef long long ll;
typedef vector<string> vs;
typedef vector<bool> vb;
typedef vector<vb> vvb;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef set<int> si;
typedef vector<si> vsi;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef pair<int, int> pi;
typedef queue<int> qi;
typedef queue<pi> qpi;
typedef pair<ll, ll> pll;
typedef queue<pll> qpll;
typedef vector<pi> vpi;
typedef vector<pll> vpll;
const int mod = 1000000007;
const int INF = 1001001001;
// 小数点　<< fixed << setprecision(0) <<
// sort降順　sort(ALL(),greater<int>());
// 円周率　M_PI
// 文字判定　isupper islower
// 文字変換　
// 順列　do {} while(next_permutation(ALL(X)));
// 最大値　LLONG_MAX
// a内でx以上　auto iter = lower_bound(ALL(a), x);
// a内でxより大きい　auto iter = upper_bound(ALL(a), x);
int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
struct edge {
    int from; //出発点
    int to;   //到達点
    int cost; //移動コスト
};
typedef struct edge se;
typedef vector<edge> ve;
unsigned Euclidean_gcd(unsigned a, unsigned b) {
  if(a < b) return Euclidean_gcd(b, a);
  unsigned r;
  while ((r=a%b)) {
    a = b;
    b = r;
  }
  return b;
}

vpll PrimeFactorization(ll n) {
  vpll res;
  for (ll i=2; i*i<=n; i++) {
    if(n%i!=0) continue;
    ll ex=0;
    while(n%i==0) {
      ex++;
      n/=i;
    }
    res.push_back(pll(i,ex));
  }
  if(n!=1) res.push_back(pll(n,1));
  return res;
}

int main() {
  ll n;
  cin >> n;
  ll i=0;
  while(n>0) {
    i++;
    n-=pow(26,i);
  }
  vll ans(i,0);
  n+=pow(26,i);
  REP(j,i) {
    ans[i-1-j]=n%26;
    n/=26;
  }
  ans[i-1]--;
  REP(j,i) {
    char s='a'+ans[j];
    cout << s;
  }
  cout << endl;
  return 0;
}