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>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> l_l;
typedef pair<int, int> i_i;
typedef vector<ll> vl;
typedef vector<int> vi;
typedef vector<char> vec;
typedef vector<bool> veb;
typedef vector<string> ves;
typedef vector<vector<ll>> vvl;
typedef vector<vector<int>> vvi;
typedef vector<vector<char>> vvec;
typedef vector<vector<bool>> vveb;
typedef vector<vector<string>> vves;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<(int)(n);i++)
#define rep2(i,n) for(int i=2;i<(int)(n);i++)
#define repk(i,k,n) for(int i=k;i<(int)(n);i++)
#define rev(v) reverse(v.begin(), v.end())
#define fs first
#define sc second
#define pb push_back
#define pp pop_back
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define maxel(a) *max_element(all(a))
#define minel(a) *min_element(all(a))
#define acc accumulate
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
#define mod (1000000007)
typedef long long int64;
//const int64 INF = 1LL << 58;
#define dame { puts("-1"); return 0;}
#define YES { cout << "YES" << endl; return 0; }
#define NO { cout << "NO" << endl; return 0; }
#define Yes { cout << "Yes" << endl; return 0; }
#define No { cout << "No" << endl; return 0; }
#define uniq(a) sort(all(a));(a).resize(unique(all(a)) - a.begin());
#define dab(v) for(auto i:v) cout << i << ' ' << endl;
#define vin(v) for (ll i = 0; i < v.size(); ++i) cin >> v[i];
#define vin2(v, w) for (ll i = 0; i < v.size(); ++i) cin >> v[i] >> w[i];
template<class T> bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T& a, const T& b) { if (a > b) { a = b; return 1; } return 0; }
int dx[] = { 0, 0, -1, 1 };
int dy[] = { 1, -1, 0, 0 };
int gcd(int a, int b) { if (a % b == 0) { return(b); } else { return(gcd(b, a % b)); } }


int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	//cout << fixed << setprecision(18);

	long long int N;
	long long int temp_N;
	cin >> N;

	vi a;
	while (double(N) / double(26) > 0) {
		temp_N = (long long int)((N-1) / 26);
		//cout << N << " " << temp_N << endl;
		a.push_back((N-1) % 26);
		N = temp_N;
	}
	reverse(all(a));
	rep(i, a.size()) {
		cout << char('a' + a.at(i));
	}
	return 0;
}