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>
#define ll long long
#define sz(x) (int)(x).size()
using namespace std;
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
//uniform_int_distribution<int>(1000,10000)(rng)

ll binpow(ll a, ll b)
{
    if (b == 0)
        return 1;
    ll res = binpow(a, b / 2);
    res*=res;
    if (b % 2)
        return res * a;
    return res;
}

ll gcd(ll a,ll b)
{
    if (b==0) return a;
    return gcd(b,a%b);
}

string to_upper(string a)
{
    for (int i=0;i<(int)a.size();++i) if (a[i]>='a' && a[i]<='z') a[i]-='a'-'A';
    return a;
}
 
string to_lower(string a)
{
    for (int i=0;i<(int)a.size();++i) if (a[i]>='A' && a[i]<='Z') a[i]+='a'-'A';
    return a;
}

int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    string s;
    int n=0;
    cin>>s;
    for (auto c : s)
        n+=(c-'0');
    if (n%9)
        cout<<"No";
    else
        cout<<"Yes";
    return 0;
}