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 (Clang 10.0.0) (judge verdict: Runtime Error)

#include <stdio.h>

unsigned int numberDigits( unsigned long long int n )
{
    const unsigned int Base    = 10;

    return 1 +  ( ( n /= Base ) == 0 ? 0 : numberDigits( n ) );
}

int main(void) 
{
    enum { N = 1000 };
    char s[N + 1];

    while ( scanf( "%s", s ) == 1 && !( s[0] == '0' && s[1] == '\0' ) )
    {
        const unsigned int Divisor = 9;

        unsigned int sum = 0;

        for ( size_t i = 0; s[i] != '\0'; i++ )
        {
            sum += s[i] - '0'; 
        }


        if ( sum != 0 && sum % Divisor == 0 )
        {
            printf( "%s is a multiple of %u and has %u-degree %u.\n", 
                    s, Divisor, Divisor, numberDigits( sum ) );
        }
        else
        {
            printf( "%s is not a multiple of %u.\n", s, Divisor );
        }
    }

    return 0;
}