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<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef pair<int,int> pii;
typedef pair<double, double> pdd;
typedef pair<ll, ll> pll;
typedef vector<pii> vii;
typedef vector<pll> vll;
typedef vector<pdd> vdd;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define mem(a,b) memset(a, b, sizeof(a) )
#define all(x) (x).begin(),(x).end()
#define INF 1000000000000
#define MOD 10000007
#define PB push_back
#define MP make_pair
#define F first
#define S second

inline void normal(ll &a) { a %= MOD; (a < 0) && (a += MOD); }
inline ll modMul(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a*b)%MOD; }
inline ll modAdd(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a+b)%MOD; }
inline ll modSub(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while(p) { if(p&1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, MOD-2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a*b/gcd(a,b);}

int main(){
    int a,b;
    cin>>a>>b;
    if(abs(a-b)%2==0){
        cout<<abs(a-b)/2+min(a,b);
    }
    else cout<<"IMPOSSIBLE";
}