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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <iostream>
#include <string>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <cstring>

using namespace std;

#define rep(i,x) for(int i=0;i<x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define FIND(x, target) find(x.begin(),x.end(),target)
#define pb(x) push_back(x)
#define output(x) cout<<x<<endl
#define ERASE(x, n) x.erase(x.begin(), x.begin()+n)
// #define int long long
#define P pair<int, int>
typedef long long ll;

const long long int MOD = 1e9+7;
const int INF = -1;

signed main(void) {
  	int A, B;
  	cin >> A >> B;

  	if (A == B) {
  		cout << 0 << endl;
  	} else {
  		int K = (A + B)/2;
  		float k = float(A + B)/2.0;
  		if ((A-K)*(B-K) <= 0 && K == k) {
  			cout << K << endl;
  		} else {
  			cout << "IMPOSSIBLE" << endl;
  		}
  	}

	return 0;
}