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 digit_sum(int n){
	unsigned long long sum;
	if (n / 10 == 0){
		return (n);
	}
	sum = digit_sum(n / 10) + (n % 10);
	return (sum);
}
int main(void)
{
	unsigned long long N, ans;

	if (scanf("%llud", &N) < 1){
		return (0);
	}
	ans = digit_sum(N);
	printf("%llud\n", ans);
	if  (ans % 9 == 0){
		printf("yes\n");
	}else{
		printf("no\n");
	}
	return (0);
}