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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

public class Main {

    static void solve() {
        int a = nextInt();
        int b = nextInt();

        if (a % 3 == 0 || b % 3 == 0 || (a + b) % 3 == 0) {
            out.println("Possible");
        } else {
            out.println("Impossible");
        }
    }

    static final int MOD = 1_000_000_007;
    static long[] fac, finv, inv;

    // nCkの初期化
    static void comInit(int max) {
        fac = new long[max];
        finv = new long[max];
        inv = new long[max];
        fac[0] = fac[1] = 1;
        finv[0] = finv[1] = 1;
        inv[1] = 1;

        for (int i = 2; i < max; i++) {
            fac[i] = fac[i - 1] * i % MOD;
            inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
            finv[i] = finv[i - 1] * inv[i] % MOD;
        }
    }

    // nCkの計算
    static long com(int n, int k) {
        if (n < k)
            return 0;
        if (n < 0 || k < 0)
            return 0;
        return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
    }

    static PrintWriter out;
    static Scanner sc;

    static int[][] newIntArray(int h, int w, int value) {
        int[][] ret = new int[h][w];
        for (int i = 0; i < h; i++) {
            for (int j = 0; j < w; j++) {
                ret[i][j] = value;
            }
        }
        return ret;
    }

    static int nextInt() {
        return Integer.parseInt(sc.next());
    }

    static long nextLong() {
        return Long.parseLong(sc.next());
    }

    static String nextString() {
        return sc.next();
    }

    static int[] nextIntArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        return a;
    }

    static List<Integer> nextIntList(int n) {
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(nextInt());
        }
        return list;
    }

    static List<Double> nextDoubleList(int n) {
        List<Double> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add((double) nextInt());
        }
        return list;
    }

    static List<Long> nextLongList(int n) {
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(nextLong());
        }
        return list;
    }

    static char[][] nextCharArray(int h, int w) {
        char[][] c = new char[h][w];
        for (int i = 0; i < h; i++) {
            String str = nextString();
            for (int j = 0; j < w; j++) {
                c[i][j] = str.charAt(j);
            }
        }
        return c;
    }

    static <T extends Comparable<? super T>> void sort(List<T> list) {
        Collections.sort(list);
    }

    // greatest common divisor
    // 最大公約数
    static long gcd(long a, long b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }

    // least common multiple
    // 最小公倍数
    static long lcm(long a, long b) {
        if (a >= b && a % b == 0)
            return a;
        if (b > a && b % a == 0)
            return b;

        if (a > b) {
            return (a / gcd(a, b)) * b;
        } else {
            return (b / gcd(a, b)) * a;
        }
//      return a * b / gcd(a, b);
    }

    // baseのn乗を計算を返す
    static int pow(int base, int n) {
        int ret = 1;
        for (int i = 0; i < n; i++) {
            ret *= base;
        }
        return ret;
    }

    // return n^k mod m
    static long powMod(long n, long k, long m) {
        if (k == 0) {
            return 1;
        } else if (k % 2 == 1) {
            return powMod(n, k - 1, m) * n % m;
        } else {
            long tmp = powMod(n, k / 2, m);
            return tmp * tmp % m;
        }
    }

    // intをlength桁のbit文字列に変換
    static String toBitString(int length, int n) {
        StringBuilder sb = new StringBuilder();
        for (int i = length - 1; i >= 0; i--) {
            if ((n >> i) % 2 == 1) {
                sb.append("1");
            } else {
                sb.append("0");
            }
        }
        return sb.toString();
    }

    public static void main(String[] args) {
        out = new PrintWriter(System.out);
        sc = new Scanner(System.in);

        solve();

        out.flush();
        sc.close();
    }

}