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<stdio.h>
#include<math.h>
#include <limits.h>
/* 値を入れ替える関数 */
void swap(int* x, int* y) {
	int temp;    // 値を一時保存する変数

	temp = *x;
	*x = *y;
	*y = temp;
}

/***
* pivotを決め、
* 全データをpivotを境目に振り分け、
* pivotの添え字を返す
***/
int partition(int array[], int left, int right) {
	int i, j, pivot;
	i = left;
	j = right + 1;
	pivot = left;   // 先頭要素をpivotとする

	do {
		do { i++; } while (array[i] < array[pivot]);
		do { j--; } while (array[pivot] < array[j]);
		// pivotより小さいものを左へ、大きいものを右へ
		if (i < j) { swap(&array[i], &array[j]); }
	} while (i < j);

	swap(&array[pivot], &array[j]);   //pivotを更新

	return j;
}

/* クイックソート */
void quick_sort(int array[], int left, int right) {
	int pivot;

	if (left < right) {
		pivot = partition(array, left, right);
		quick_sort(array, left, pivot - 1);   // pivotを境に再帰的にクイックソート
		quick_sort(array, pivot + 1, right);
	}
}

int main() {
	long int n;
	int i;
	char s[1000];
	scanf("%ld", &n);

	for (i = 0; n > 0; i++) {
		s[i] = 'a' + (n % 26 - 1);
		n = n / 26;
	}
	while (i > 0) {
		i--;
		printf("%c", s[i]);
	}
	printf("\n");
	return 0;
}