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: Wrong Answer)

#include "bits/stdc++.h"
using namespace std;
using ll=long long;
#define print(n) cout<<n<<endl
#define rep(i,a,n) for(ll i=a;i<n;i++)
#define ALL(s) s.begin(),s.end()
#define P pair<ll,ll>
#define vl vector<ll>
#define vi vector<int>
#define vvl vector<vector<ll>>
#define vvi vector<vector<int>>
const int M=200111;
const int inf=1000000007;
const int MOD=1000000007;
const double PI = 3.1415926535897932384626433;
const long long INF=1e18;
int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};
int ddx[8]={1,-1,0,0,1,-1,-1,1},ddy[8]={0,0,1,-1,1,-1,1,-1};


struct UnionFind {
	vector<int> par;
	UnionFind(int n) : par(n, -1) { }

	int root(int x) {
		if (par[x] < 0) return x;
		else return par[x] = root(par[x]);
	}

	bool same(int x, int y) {
		return root(x) == root(y);
	}

	bool merge(int x, int y) {
		x = root(x); y = root(y);
		if (x == y) return false;
		if (par[x] > par[y]) swap(x, y); // merge technique
		par[x] += par[y];
		par[y] = x;
		return true;
	}

	int size(int x) {
		return -par[root(x)];
	}
};

vvl comb(int n) {
	vvl v(n + 1,vl(n + 1, 0));
	for (int i = 0; i < v.size(); i++) {
		v[i][0] = 1;
		v[i][i] = 1;
	}
	for (int j = 1; j < v.size(); j++) {
		for (int k = 1; k < j; k++) {
			v[j][k] = (v[j - 1][k - 1] + v[j - 1][k]);
		}
	}
	return v;
}

void sosu(ll x,bool f[]){
	ll y=x;
	for(ll i=2;i<=sqrt(y);i++){
		if(x%i==0){
			x/=i;
			f[i]=true;
			i--;
		}
	}
}

ll __gcd(ll x,ll y){
	if(x<y)swap(x,y);
	while(1){
		ll tmp=x%y;
		x=y;
		y=tmp;
		if(y==0)break;
	}
	return x;
}

int keta(ll a){
	int ke=1;
	while(a/=10)ke++;
	return ke;
}

vector<long long> enum_divisors(long long N) {
	vector<long long> res;
	for (long long i = 1; i * i <= N; ++i) {
		if (N % i == 0) {
			res.push_back(i);
			// 重複しないならば i の相方である N/i も push
			if (N/i != i) res.push_back(N/i);
		}
	}
	// 小さい順に並び替える
	sort(res.begin(), res.end());
	return res;
}

ll zyou(ll x,ll n){
	ll y=1;
	rep(i,0,n){
		y*=x;
	}
	return y;
}

int main() {
	int a,b;
	cin>>a>>b;
	bool f=false;
	if(a%3==0)f=true;
	if(b%3==0)f=true;
	if(a+b%3==0)f=true;
	if(f)print("Possible");
	else print("Impossible");
}