Problem:
Score : 700 points

Problem Statement

Aoki loves numerical sequences and trees.

One day, Takahashi gave him an integer sequence of length N, a_1, a_2, ..., a_N, which made him want to construct a tree.

Aoki wants to construct a tree with N vertices numbered 1 through N, such that for each i = 1,2,...,N, the distance between vertex i and the farthest vertex from it is a_i, assuming that the length of each edge is 1.

Determine whether such a tree exists.

Constraints

2 ≦ N ≦ 100

1 ≦ a_i ≦ N-1

Input

The input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

If there exists a tree that satisfies the condition, print Possible. Otherwise, print Impossible.

Sample Input 1

5
3 2 2 3 3

Sample Output 1

Possible

The diagram above shows an example of a tree that satisfies the conditions. The red arrows show paths from each vertex to the farthest vertex from it.

Sample Input 2

3
1 1 2

Sample Output 2

Impossible

Sample Input 3

10
1 2 2 2 2 2 2 2 2 2

Sample Output 3

Possible

Sample Input 4

10
1 1 2 2 2 2 2 2 2 2

Sample Output 4

Impossible

Sample Input 5

6
1 1 1 1 1 5

Sample Output 5

Impossible

Sample Input 6

5
4 3 2 3 4

Sample Output 6

Possible

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <sstream>
#include <algorithm>
#include <vector>
#include <utility>
#include <functional>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <deque>
#include <ctime>
  
using namespace std;
  
#define rep(i,n) REP(i,0,n)
#define REP(i,s,e) for(int i=(s); i<(int)(e); i++)
#define pb push_back
#define mp make_pair
#define all(r) r.begin(),r.end()
#define rall(r) r.rbegin(),r.rend()
#define fi first
#define se second
#define println(X) cout<<X<<endl;
#define DBG(X) cout<<#X<<" : "<<X<<endl;
  
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
 
const int INF = 1e9;
const ll MOD = 1e9 + 7;
double EPS = 1e-8;
const double PI = acos(-1);

int main(){
	int n;
	cin>>n;
	vi a(n);
	rep(i, n) cin>>a[i];
	sort(all(a));
	map<int,int> m;
	rep(i, n) m[a[i]]++;
	bool f = true;
	//if(m[a.back()] < 2) f = false;
	if(a.back()%2==0){
		if(m[a[0]] == 1);
		else f = false;
	}
	else{
		if(m[a[0]]==2);
		else f = false;
	}
	for(auto& p : m){
		if(p.fi==a[0]) continue;
		else if(p.se<2) f = false;
	}
	cout<<(f?"Possible":"Impossible")<<endl;

}