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>
#include <string.h>
#include <ctype.h>
#include <stdlib.h>
#include <unistd.h>

int main()
{
	char *s;
	int buff = 5;
	int ans;
	int ret;

	ans = 0;
	s = malloc((buff + 1) * sizeof(char));
	while ((ret = read(0, s, buff))) {
		if (ret == -1)
			printf("fail\n");
		s[ret] = '\0';
		for (int i = 0; i < ret; i++) {
			if (isdigit(s[i]))
				ans = (ans + (s[i] - '0') % 9) % 9;
		}
		if (strchr(s, '\n'))
			break;
	}
	if (ans % 9 == 0)
		printf("Yes\n");
	else
		printf("No\n");
	return (0);
}