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: Runtime Error)

#include <stdio.h>

int main(void)
{
	long long int N, ex_N;
	int remain; 	//余り
	int number[10000], rev_number[10000];	//26で割った余りを入れる
	int flag;
	int i;
	char s[26] = {'a','b','c','d','e','f','g','h','i','j','k','l','m',
					'n','o','p','q','r','s','t','u','v','w','x','y','z'};

	scanf("%lld", &N);

	if (N / 26 == 0) {
		printf("%c\n", s[(N % 26) - 1]);
	}else {
		//26で割れる回数を数える
		ex_N = N;
		while (ex_N != 0) {
			ex_N /= 26;
			flag += 1;
		}

		//26で割った余りを配列に入れる
		for (i=0; i<flag; i++) {
			remain = N % 26;
			number[i] = remain;
			N /= 26;
		}
		//あまりの順序を逆にする
		for (i=0; i<flag; i++) {
	        rev_number[i] = number[flag-1-i];
	    }

		for (i=0; i<flag; i++) {
			printf("%c", s[rev_number[i]-1]);
		}
		printf("\n");
	}	
}