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: Wrong Answer)


 
int main()
{
 
  /* 自然数 */
  unsigned long long int num;
 
  /* 各桁の数字 */
  unsigned long long int dig;
 
  /* 和 */
  unsigned long long int sum = 0;
 
  //printf("自然数を入力してください = ");
    scanf("%lld", &num);
 
  /* 自然数の各桁の数字の和を計算 */
  while(num){
    dig = num % 10;
    sum = sum + dig;
    num = num / 10;
  }
 
  /* 各桁の数字の和を出力 */
  //printf("自然数の各桁の数字の和 = %llu\n", sum);
 
    
//    if(sum % 9){
//      puts("No\n");
//    }else{
//      puts("Yes\n");
//    }
    if(sum % 9==0){
      puts("Yes\n");
    }else{
      puts("No\n");
    }
  
    
  return 0;
}