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>
int main(){
	int N, i = 0, base = 1, sum = 0;
	int k = 0, ssum;
	scanf("%d", &N);
	int n = N;
	while(n != 0){
		n /= 10;
	i++;
	}
  	while(k != i){
		base *= 10;
		k++;
	}
	base /= 10;
	while(base > 0){
		ssum = N / base;
		sum += ssum;
		N %= base;
		base /= 10;
	}
	int nine = sum % 9; 
	if(nine == 0){
    	printf("Yes\n");
    }else{
      	printf("No\n");
    }
	return 0;
}