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 <bits/stdc++.h>
#include <cstdlib>  //abs()で絶対値を取得する（整数）
#include <cmath>    //abs()かfabs()で絶対値を取得する（少数）
#define rep(i,n) for (int i = 0; i < (int)(n); i++)
using namespace std;
using ll = long long;
int main(){
    ll n;
    ll a = 0;
    ll b = 1;
    ll c = 0;
    a = log10(n) + 1;
    for (int i = 0; i < a; i++ ){
        for (int q = 0; q < i; q++){
            b = b * 10;
        }
        c = c + n % b;
    }
    if (c % 9 == 0){
        cout << "Yes";
    }else{
        cout << "No";
    }
}