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 <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define REP(i, j, n) for (int i = j; i < (int)(n); i++)
#define brep(i, n) for (int i = n; i > 0; i--)
#define BREP(i, j, n) for(int i = n; i > j; i--)
#define TEST cout << "PASS" << endl
#define all(x) (x).begin(),(x).end()
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1e9 + 1e6 + 1e3 + 1;
const ll MOD = 1e9 + 7;
const double PI = 3.141592653589793;

int main(){
    string s;
    cin >> s;
    int cnt = 0;
    for(int i = 0; i < int(s.size()); i++){
        int k = s[i] - 48;
        cnt += k;
        cnt %= 9;
    }
    if(cnt == 0) cout << "Yes" <<endl;
    else cout << "No" << endl;
}