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: Accepted)

#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
#include<numeric>
using namespace std;

typedef long long ll;
typedef unsigned long long ul;
typedef unsigned int ui;
typedef pair<int, int>P;
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
typedef vector<vector<ll>> mat;
typedef vector<int> vec;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
const ll mod = 1000000007;
const ll INF = mod * mod;
const ld eps = 1e-12;
const ld pi = acos(-1.0);
//int qp(int a,ll b){int ans=1;do{if(b&1)ans=1ll*ans*a%mo;a=1ll*a*a%mo;}while(b>>=1);return ans;}
ll qp(ll a,ll b,int mo){int ans=1;do{if(b&1)ans=1ll*ans*a%mo;a=1ll*a*a%mo;}while(b>>=1);return ans;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
int dx[4] = { 0,1,0,-1 };
int dy[4] = { 1,0,-1,0 };

void solve() {
    ll a, b, c, n, k, x;
    string s, t;
    ll ans = 0;
    double dans = 0;
    string sans = "";



    cin  >> a >> b;
    ll length = max(a,b) - min(a,b);
    if(length%2 == 1) cout << "IMPOSSIBLE";
    else {
        ll ans1 = max(a,b) - length/2;
        ll ans2 = max(a,b) + length/2;

        ll tmp1 = max(min(a,b) - ans1, ans1 - min(a,b));
        ll tmp2 = max(min(a,b) - ans2, ans2 - min(a,b));

        if(tmp1 == length/2) std::cout << ans1 << endl; 
        else if(tmp2 == length/2) std::cout << ans2 << endl; 
        else  cout << "IMPOSSIBLE";
    }




    // std::cout << ans << endl; 
    // std::cout << sans << endl; 
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(10);
    solve();
    // stop
        return 0;
}