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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define rng(a) a.begin(),a.end()
#define rrng(a) a.rbegin(),a.rend()  // 右から読む.reverse
#define sortu(x) sort((x).begin(),(x).end())
#define sortd(x) sort((x).begin(),(x).end(), greater<int>())
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
//#define pcnt __builtin_popcountll 
#define uni(x) x.erase(unique(rng(x)),x.end())
//#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define show(x) cerr<<#x<<" = "<<x<<endl;
//#define PQ(T) priority_queue<T,v(T),greater<T> >
//#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define newline puts("")
#define v(T) vector<T>
#define vv(T) v(v(T))
using namespace std;
using ll = long long int;
//typedef unsigned uint;
//typedef unsigned long long ull;
typedef pair<int,int> P;
typedef tuple<int,int,int> T;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<P> vp;  // 2つ
typedef vector<T> vt;  // 3つ
//int getInt(){int x;scanf("%d",&x);return x;}
template<typename T>istream& operator>>(istream&i,v(T)&v){rep(j,sz(v))i>>v[j];return i;}  // cin>>v 配列に代入
template<typename T>ostream& operator<<(ostream&o,const v(T)&v){if(sz(v))o<<join(v);return o;}  // cout<<v 配列を出力
//template<typename T>string join(const v(T)&v){stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T1,typename T2>istream& operator>>(istream&i,pair<T1,T2>&v){return i>>v.fi>>v.se;}
template<typename T1,typename T2>ostream& operator<<(ostream&o,const pair<T1,T2>&v){return o<<v.fi<<","<<v.se;}  // ,に注意
template<typename T>bool mins(T& x,const T&y){if(x>y){x=y;return true;}else return false;}
template<typename T>bool maxs(T& x,const T&y){if(x<y){x=y;return true;}else return false;}
template<typename T>ll suma(const v(T)&a){ll res(0);for(auto&&x:a)res+=x;return res;}
//const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
const int MOD = 1e9+7;
#define dame0 { puts("0"); return 0;}
#define dame1 { puts("-1"); return 0;}
#define yn {puts("Yes");}else{puts("No");}
#define yn2 {puts("YES");}else{puts("NO");}
#define yn3 {puts("yes");}else{puts("no");}
#define on {puts("ok");}else{puts("ng");}
#define on2 {puts("OK");}else{puts("NG");}
//#define perm(c) sort(all(c));for(bool c##p=1;c##p;c##p=next_permutation(all(c)))
#define outa {cout << ans << endl;}
#define out(x) {cout << x << endl;}
using Graph = vector<vector<int>>;
//#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define posi(x,v) *((lower_bound(x.begin(),x.end(),v)))
#define posl(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())  // v未満の個数. keyを前に挿入(iterはkeyより大)
#define posl2(x,v) (x.end()-lower_bound(x.begin(),x.end(),v))  // v以上の個数
#define posu(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())  // v以下の個数　keyを後ろに挿入
#define posu2(x,v) (x.end()-upper_bound(x.begin(),x.end(),v))  // vより大きい個数
#define bs(x,v) (binary_search(x.begin(),x.end(),v))
//mins(ans,now);  ansに最小値が入る
//if(max(ans,now)) ansが小さければ更新して実行
//ll sum = suma(a);
//int a[33][4];
//vi a{1,2,3};
// #include <set> 重複するデータを保持しない
// set<int> st;
// st.insert(a);
// st.size() // 配列の長さ
// st.count(x) xの数を返す
// st.find(x) xのイテレータを返す
// for (auto x : st) 要素の取得。

bool flg = false;

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

 /*
g++ a.cpp
./a.out

oj d
oj t

debug:ctrl+shift+B problemin

*/