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 a[200000],sum = 0,count = 0;
    scanf("%d",&a[200000]);
    sum = a[200000];
    if(a[200000] == 0){
        printf("Yes\n");
    }
    else{
    while(sum != 0){
        sum = sum%10;
        count+=sum;
    if(count%9 == 0){
        printf("Yes\n");
        break;
    }
    else{
        printf("No\n");
        break;
    }
    }
    }
    return 0;
}