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: Accepted)

#include <stdio.h>
 
int main(void) {
  char n[3000000];
  int i=0;
  int sum = 0;
  
  fgets(n,sizeof(n),stdin);
  while (n[i] != '\0') {
    if ('0'<=n[i] && n[i]<='9') {
      sum += n[i] - '0';
    }
    i++;
  }
  printf("%s", sum%9==0 ? "Yes" : "No");
  return 0;
}