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: Runtime Error)

#include<stdio.h>
int main(){
  int unused __attribute__((unused));
  int i, sum = 0;
  char N[200];
  unused = scanf("%s", N);

  i = 0;
  while(N[i]!='\0'){
    sum += (int)N[i]-97;
    i++;
  }
  if(sum%9 == 0 || sum==0) printf("Yes");
  else printf("No");
  return 0;
}