Problem:
Score : 100 points

Problem Statement

Snuke is giving cookies to his three goats.

He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins).

Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.

Constraints

1 \leq A,B \leq 100

Both A and B are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

If it is possible to give cookies so that each of the three goats can have the same number of cookies, print Possible; otherwise, print Impossible.

Sample Input 1

4 5

Sample Output 1

Possible

If Snuke gives nine cookies, each of the three goats can have three cookies.

Sample Input 2

1 1

Sample Output 2

Impossible

Since there are only two cookies, the three goats cannot have the same number of cookies no matter what Snuke gives to them.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using ll = long long;
using namespace std;
#define rep(i,n) for(int i=0, i##_len=(int)(n); i<i##_len; i++)
#define reps(i,n) for(int i=1 , i##_len=(int)(n);i<=i##_len;i++)
#define rrep(i,n) for(int i=((int)(n)-1);i>=0;i--)
#define rreps(i,n) for(int i=((int)(n));i>0;i--)
#define repi(i,x) for(auto i=(x).begin(),i##_fin=(x).end();i!=i##_fin;i++)
#define all(x) (x).begin(), (x).end()
#define F first
#define S second
#define mp make_pair
#define pb push_back
#define solve(a) ((a)?"Possible":"Impossible")
typedef vector<int> Vi;
typedef vector<Vi> VVi;
typedef pair<int , int> Pi;
typedef vector<Pi> VPi;
typedef vector<long long> V;
typedef vector<V> VV;
typedef pair<long long , long long> P;
typedef vector<P> VP;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1;} return 0;}
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1;} return 0;}
const long long INFLL = 1LL<<60;
const int INF = 1<<30;
const double PI=acos(-1);
int main(){
	int a,b;
	bool ans=false;
	cin >>a>>b;

	cout<<solve(a%3==0 or b%3==0 or (a+b)%3==0)<<endl;
}