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<stdlib.h>
#include<string.h>
#include<math.h>
#define rep(i,n) for(int i=0;i<n;i++)
int gcd(int a, int b){
    int r=1;
    while(r!=0){
        r=a%b;
        a=b;
        b=r;
    }
    return a;
}
void revStr(char* str){
    int size = strlen(str);
    int i,j;
    char tmp = {0};
    for(i = 0, j = size - 1; i < size / 2; i++, j--){
        tmp = str[i];
        str[i] = str[j];
        str[j] = tmp;
    }return;
}
typedef struct{
    int a,b,c;
}data;

//昇順
int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
}
//構造体降順
int comp( const void *c1, const void *c2 )
{
  data test1 = *(data *)c1;
  data test2 = *(data *)c2;
  int tmp1 = test1.b;   /* b を基準とする */
  int tmp2 = test2.b;
  return tmp2 - tmp1;
}


int kai(int n){
    if(n==0){
        return 1;
    }
    int sum=1;
    for(int i=1;i<=n;i++){
        sum*=i;
    }
    return sum;
}

int main(){
    long long N;
    scanf("%lld",&N);
    long long temp=0;
    temp=N;
    long long p=1;
    while(temp>0){
        temp-=(long long)pow(26,p);
        p++;
    }
    long long sum=0;
    long long q=1;
    while(q<p-1){
        sum+=(long long)pow(26,q);
        q++;
    }
    char vl[p-1];
    rep(i,p-1){
        vl[i]='a';
    }
    for(long long i=1;i<N-sum;i++){
        for(long long j=0;j<=(i/27);j++){
            vl[p-j-2]+=i%27;
        }
    }
    printf("%s",vl);
}