Codeforces 344B Simple Molecules

Description

Mad scientist Mike is busy carrying out experiments in chemistry. Today he will attempt to join three atoms into one molecule.

A molecule consists of atoms, with some pairs of atoms connected by atomic bonds. Each atom has a valence number — the number of bonds the atom must form with other atoms. An atom can form one or multiple bonds with any other atom, but it cannot form a bond with itself. The number of bonds of an atom in the molecule must be equal to its valence number.

Mike knows valence numbers of the three atoms. Find a molecule that can be built from these atoms according to the stated rules, or determine that it is impossible.

Input

The single line of the input contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 10⁶) — the valence numbers of the given atoms.

Output

If such a molecule can be built, print three space-separated integers — the number of bonds between the 1-st and the 2-nd, the 2-nd and the 3-rd, the 3-rd and the 1-st atoms, correspondingly. If there are multiple solutions, output any of them. If there is no solution, print "Impossible" (without the quotes).

Sample Input

Input

1 1 2

Output

0 1 1

Input

3 4 5

Output

1 3 2

Input

4 1 1

Output

Impossible

题意：这道题告诉你3个节点的度数，也就是相邻两点的边的数目，例如：我们定义这三个顶点分别是a,b,c，设第一个点和第二个点之间的边的数目为x,第二个点与第三个点之间的边的数目为y,第三个点与第四个
点之间的边的数目为z,那么a = x + z, b = x + y, c = y + z;而且根据握手定理： 顶点度数之和为变得2倍，所以顶点度数和为偶数，而且两点见得边数不小于0.  有个疑点是为什么x = a +b - c?

/*************************************************************************
    > File Name: cf.cpp
    > Author: PrayG
    > Mail: 
    > Created Time: 2016年07月10日 星期日 12时57分34秒
 ************************************************************************/

#include<iostream>
#include<cstdio>
#include<bits/stdc++.h>
#include<string>
using namespace std;
const int maxn = 200005;
int main()
{
    int a,b,c;
    cin >> a >> b >> c;
    int x ,y,z;
    int sum;
    sum = a+ b + c;
    x = a + b - c;
    y = b + c - a;
    z = a + c - b;
    if(x < 0 || y < 0 || z < 0|| sum%2)
    {
        printf("Impossible\n");
    }
    else
    {
        printf("%d %d %d\n",x/2,y/2,z/2);
    }
    return 0;
}