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>

int main(void){
    char cstr[200001];

    long long int sum;
    int i;
    (void)scanf("%s", &cstr);

    sum = 0;
    for (i =0;i<200001;++i){
        if (cstr[i] == '\0') {
            break;
        }
        sum += (int)cstr[i] - 48;
        // printf("%d\n", (int)cstr[i] - 48);
    }
    if (sum % 9 == 0) {
        printf("Yes\n");
    } else {
        printf("No\n");
    }

}