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)

/*
 * じょえチャンネル
 * 高評価・チャンネル登録よろしくおねがいします！
 * https://www.youtube.com/channel/UCRXsI3FL_kvaVL9zoolBfbQ
 */

#include <bits/stdc++.h>

//#pragma GCC target("avx2")
//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")

#define f(i,n) for(int i=0;i<(n);i++)


#define inf (int)(3e18)

//here!!!
// define int long long !!!!!

#define int long long

// define int long long !!!!!


#define mod (int)((1e9)+7)
//constexpr int mod = 998244353ll;

#define intt long long
#define itn long long
#define P pair<int,int>

#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)

#define ALL(v) v.begin(),v.end()

using namespace std;

//Library
//モッドパウ
inline int modpow(int x, int y, int m = mod) {
    int res = 1;
    while (y) {
        if (y % 2) {
            res *= x;
            res %= m;
        }
        x = x * x % m;
        y /= 2;
    }
    return res;
}

int mypow(int x, int y) {
    int res = 1;
    while (y) {
        if (y % 2) {
            res *= x;
        }
        x = x * x;
        y /= 2;
    }
    return res;
}
//is the number (x) a prime number?
bool prime(int x) {
    for (int i = 2; i * i <= x; i++) {
        if (!(x % i)) {
            return false;
        }
    }
    return true;
}


//saidai-kouyakusuu
inline int gcd(int x, int y) {
    if (!y) {
        return x;
    }
    return gcd(y, x % y);
}


//number of keta
int keta(int x) {
    int ans = 0;
    while (x) {
        x /= 10;
        ans++;
    }
    return ans;
}

//sum of keta
int ketasum(int x) {
    int ans = 0;
    while (x) {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}

inline int lcm(int x, int y) {
    int ans = x / gcd(x, y) * y;
    return ans;
}
int twobeki(int x) {
    int ans = 0;
    while (1) {
        if (!(x & 1)) {
            ans++;
            x /= 2;
        }
        else {
            break;
        }
    }
    return ans;
}


template <class T, class U>
inline bool chmax(T& lhs, const U& rhs) {
    if (lhs < rhs) {
        lhs = rhs;
        return 1;
    }
    return 0;
}
template <class T, class U>
inline bool chmin(T& lhs, const U& rhs) {
    if (lhs > rhs) {
        lhs = rhs;
        return 1;
    }
    return 0;
}
void Yes(){
    cout<<"Yes"<<endl;
}
void No(){
    cout<<"No"<<endl;
}
void YES(){
    cout<<"YES"<<endl;
}
void NO(){
    cout<<"NO"<<endl;
}


#define fin(i) scanf("%lld",&i)
#define fout(i) printf("%lld",i)
#define fendl printf("\n")

int kai(int x, int y) {
    int res = 1;
    for (int i = x - y + 1; i <= x; i++) {
        res *= i; res %= mod;
    }
    return res;
}

int comb(int x, int y) {
    if (y > x)return 0;
    //    cout<<kai(x, y)<<' '<<modpow(kai(y, y), mod - 2)<<endl;
    return kai(x, y) * modpow(kai(y, y), mod - 2) % mod;
}
//Library-End


#define vecin(v) for(int i=0;i<v.size();i++)scanf("%lld",&v[i]);
#define vecout(v) {for(int i=0;i<(int)v.size();i++)printf("%lld ",v[i]);printf("\n");}

//SegTree
template <class T>
class SegTree {
    int n;                       // 葉の数
    vector<T> data;              // データを格納するvector
    T def;                       // 初期値かつ単位元
    function<T(T, T)> operation; // 区間クエリで使う処理
    function<T(T, T)> update;    // 点更新で使う処理
    
    // 区間[a,b)の総和。ノードk=[l,r)に着目している。
    T _query(int a, int b, int k, int l, int r) {
        if (r <= a || b <= l) return def; // 交差しない
        if (a <= l && r <= b)
            return data[k]; // a,l,r,bの順で完全に含まれる
        else {
            T c1 = _query(a, b, 2 * k + 1, l, (l + r) / 2); // 左の子
            T c2 = _query(a, b, 2 * k + 2, (l + r) / 2, r); // 右の子
            return operation(c1, c2);
        }
    }
    
public:
    // _n:必要サイズ, _def:初期値かつ単位元, _operation:クエリ関数,
    // _update:更新関数
    SegTree(size_t _n, T _def, function<T(T, T)> _operation,
            function<T(T, T)> _update)
    : def(_def), operation(_operation), update(_update) {
        n = 1;
        while (n < _n) {
            n *= 2;
        }
        data = vector<T>(2 * n - 1, def);
    }
    
    // 場所i(0-indexed)の値をxで更新
    void change(int i, T x) {
        i += n - 1;
        data[i] = update(data[i], x);
        while (i > 0) {
            i = (i - 1) / 2;
            data[i] = operation(data[i * 2 + 1], data[i * 2 + 2]);
        }
    }
    
    // [a, b)の区間クエリを実行
    T query(int a, int b) {
        return _query(a, b, 0, 0, n);
    }
    
    // 添字でアクセス
    T operator[](int i) {
        return data[i + n - 1];
    }
};

#define R_MIN ([](long long a, long long b) { return min(a, b); })
#define R_MAX ([](long long a, long long b) { return max(a, b); })
#define R_SUM ([](long long a, long long b) { return a + b; })

#define NORMAL_UPDATE ([](long long a, long long b) { return b; })
#define ADD_UPDATE ([](long long a, long long b) { return a + b; })
#define MINUS_UPDATE ([](long long a, long long b) { return a - b; }

class Union_Find {
    vector<int> par;
    vector<int> rankmy;
    vector<int> ookisa;
    
public:
    Union_Find(int size) {
        par = vector<int>(size);
        rankmy = vector<int>(size);
        ookisa=vector<int>(size);
        for (int i = 0; i < size; i++) {
            par[i] = i;
            ookisa[i]=1;
        }
    }
    
    int find(int x) {
        if (par[x] == x) {
            return x;
        }
        return par[x] = find(par[x]);
    }
    
    void unite(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) {
            return;
        }
        if (rankmy[x] < rankmy[y]) {
            par[x] = y;
            ookisa[y]+=ookisa[x];
            ookisa[x]=0;
        }
        else {
            par[y] = x;
            ookisa[x]+=ookisa[y];
            ookisa[y]=0;
            if (rankmy[x] == rankmy[y]) {
                rankmy[x]++;
            }
        }
    }
    int size(int i){
        i=find(i);
        return ookisa[i];
    }
    bool same(int x, int y){
        return find(x) == find(y);
    }
    
};

//Union-Find-End

int perm[1000005];
void init_perm() {
    perm[0] = 1;
    REP(i, 1000000) {
        perm[i] = perm[i - 1] * i % mod;
    }
}

int nCk(int x, int y) {
    return perm[x] * modpow(perm[x - y], mod - 2) % mod * modpow(perm[y], mod - 2) % mod;
}

double kyori(pair<int, int> f, pair<int, int> s) {
    double ans = 0;
    double t = fabs(f.first - s.first);
    double y = fabs(f.second - s.second);
    ans = sqrt(t * t + y * y);
    return ans;
}

inline string stringmax(string& x,string& y){
    if (x.size()>y.size()) {
        return x;
    }
    if (x.size()<y.size()) {
        return y;
    }
    rep(i,x.size()){
        if (x[i]>y[i]) {
            return x;
        }
        if (x[i]<y[i]) {
            return y;
        }
    }
    return x;
}

#define endl "\n" //interactive の時に注意！！！



int a,b;
signed main() {
    ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    cin>>a>>b;
    if ((!((a+b)%3))||(!(a*b)%3)) {
        cout<<"Possible"<<endl;
    }else cout<<"Impossible"<<endl;
}