Problem:
Score : 200 points

Problem Statement

An integer N is a multiple of 9 if and only if the sum of the digits in the decimal representation of N is a multiple of 9.

Determine whether N is a multiple of 9.

Constraints

0 \leq N < 10^{200000}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If N is a multiple of 9, print Yes; otherwise, print No.

Sample Input 1

123456789

Sample Output 1

Yes

The sum of these digits is 1+2+3+4+5+6+7+8+9=45, which is a multiple of 9, so 123456789 is a multiple of 9.

Sample Input 2

0

Sample Output 2

Yes

Sample Input 3

31415926535897932384626433832795028841971693993751058209749445923078164062862089986280

Sample Output 3

No

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<ctype.h>
int main()
{
    long long int s, i, max, cnt=0;
    scanf("%lld", &s);
    while(s!=0)
    {
        s=s/10;
        cnt++;
    }
    char n[cnt];
    n[cnt]=s;
    for(i=0; i<strlen(n); i++)
    {
        max=max+n[i];
    }
    if(max==0)
    {
        printf("Yes\n");
    }
    else if(max%9==0)
    {
        printf("Yes\n");
    }
    else
    {
        printf("No\n");
    }
    return 0;
}