Problem:
Score: 100 points

Problem Statement

We have two distinct integers A and B.

Print the integer K such that |A - K| = |B - K|.

If such an integer does not exist, print IMPOSSIBLE instead.

Constraints

All values in input are integers.

0 \leq A,\ B \leq 10^9

A and B are distinct.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the integer K satisfying the condition.

If such an integer does not exist, print IMPOSSIBLE instead.

Sample Input 1

2 16

Sample Output 1

9

|2 - 9| = 7 and |16 - 9| = 7, so 9 satisfies the condition.

Sample Input 2

0 3

Sample Output 2

IMPOSSIBLE

No integer satisfies the condition.

Sample Input 3

998244353 99824435

Sample Output 3

549034394

Solution in C++14 (Clang 3.8.0) (judge verdict: Accepted)

#include<algorithm>//sort,二分探索,など
#include<bitset>//固定長bit集合
#include<cmath>//pow,logなど
#include<complex>//複素数
#include<deque>//両端アクセスのキュー
#include<functional>//sortのgreater
#include<iomanip>//setprecision(浮動小数点の出力の誤差)
#include<iostream>//入出力
#include<iterator>//集合演算(積集合,和集合,差集合など)
#include<map>//map(辞書)
#include<numeric>//iota(整数列の生成),gcdとlcm(c++17)
#include<queue>//キュー
#include<set>//集合
#include<stack>//スタック
#include<string>//文字列
#include<unordered_map>//イテレータあるけど順序保持しないmap
#include<unordered_set>//イテレータあるけど順序保持しないset
#include<utility>//pair
#include<vector>//可変長配列

// vectorの重複削除
// std::sort(test.begin(), test.end());
// test.erase(std::unique(test.begin(), test.end()), test.end());

using namespace std;
using ll = long long;
using P = pair<int, int>; // vector<P> ba(n); みたいな感じですぐ使える ba[i] = P(a, b); make_pairをしなくてもいい
// typedef vector<int> v;
template <class T>ostream &operator<<(ostream &o,const vector<T>&v){o<<"{";for(int i=0;i<(int)v.size();i++)o<<(i>0?", ":"")<<v[i];o<<"}";return o;}
#define debug(v) {cerr<<"\033[1;36m[debug]\033[m "<<#v<<" : "<<v<<endl;}
#define rep(i, c) for(int i = 0; i < int(c); i++)
#define ALL(x) (x).begin(), (x).end() //sort(ALL(obj))みたいな
#define SIZE(x) ((ll)(x).size()) //sizeをsize_tからllに直しておく
#define MAX(x) *max_element(ALL(x)) //最大値
#define pb push_back //
// const int MOD = 1000000007;
// string alpha="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
// printf("%.20f\n", ans); あれ
// const double PI = 3.141592653589793238462643383;

int main() {
  ll a,b; cin>>a>>b;
  if (a%2 != b%2) {
    cout << "IMPOSSIBLE" << endl;
  } else {
    cout << abs(a+b)/2 << endl;
  }
}