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

void fin(){
  cout<<"Impossible"<<endl;
  exit(0);
}

signed main(){
  int n;cin>>n;
  map<int,int> m;
  while(n--){
    int a;cin>>a;m[a]++;
  }
  int M=m.rbegin()->first;
  if(M==1){
    if(m[1]==2)cout<<"Possible"<<endl;
    else fin();
    return 0;
  }
  for(int i=M;i>M/2;i--)if(m[i]<2)fin();
  if(!(M&1))if(m[M/2]!=1)fin();
  else if(m[(M+1)/2]>2)fin();
  for(int i=(M-1)/2;i>0;i--)if(m[i])fin();
  cout<<"Possible"<<endl;
}