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 <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace  std;

#define ll long long
#define ull unsigned long long
#define f first
#define s second
typedef pair<int , int > pii;
typedef pair<ll,ll> pll;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
// find_by_order(k)  returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;
// erase,insert same as normal set

const ll MOD = 1e9+7;

void min_self(ll &a , ll b){a=min(a,b);}
void max_self(ll &a, ll b){a=max(a,b);}
ll add(ll x, ll y){return (x + y) % MOD;}
ll sub(ll x, ll y){return add(x, MOD - y);}
ll mul(ll x, ll y){return (x * 1ll * y) % MOD;}
ll binpow(ll x, ll y)
{   ll z = 1;
    while(y > 0)
    {if(y % 2 == 1) z = mul(z, x);
        x = mul(x, x);
        y /= 2;
    }
    return z;
}
ll inv(ll x){return binpow(x, MOD - 2);}
const int MXN = 1e6+10;
ll fact[MXN];
void fac_pre()
{
    fact[0]=1;
    for(ll i=1; i<MXN; i++){
        fact[i]=mul(fact[i-1],i);
    }
}
ll C(ll n, ll k)
{
    return mul(fact[n], inv(mul(fact[k], fact[n - k])));
}

ll gcd(ll a, ll b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
ll lcm(ll a, ll b){
    return (a*b)/gcd(a,b);
}
///--//

int  main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    //freopen("input.txt", "r", stdin);
    //freopen ("output.txt","w",stdout);
    /* KEEP IN MIND
     * take care when sorting a 1 based indexed vector
     * use greater<ll> for ll's
    */
    string s; cin>>s;
    ll sum=0;
    for(int i=0; i<s.length(); i++){
        sum+=s[i]-'0';
    }
    if(sum % 9 ==0) cout<<"YES"<<endl;
    else cout<<"NO"<<endl;

}