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;

typedef long long int ll;
typedef pair<int,int> ii;

const int N = 1e5 + 5;

int n,a[N];

multiset <int> s;

void fail(bool ok) {
  if (ok == true) {
    cout << "Impossible\n";
    exit(0);
  }
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL); cout.tie(NULL);
  cin >> n;
  for (int i = 1 ; i <= n ; i++) {
    cin >> a[i];
    s.insert(-a[i]);
  }
  while(s.empty() == false) {
    int x = -*s.begin();
    // cerr << x << endl;
    if (x % 2 == 0) {
      for (int i = x ; i >= x / 2 ; i--) {
        auto it = s.find(-i);
        fail(it == s.end());
        s.erase(it);
        if (i != x / 2) {
          auto it = s.find(-i);
          fail(it == s.end());
          s.erase(it);
        }
      }     
    }
    else {
      for (int i = x ; i > x / 2 ; i--) {
        auto it = s.find(-i);
        fail(it == s.end());
        s.erase(it);
        it = s.find(-i);
        fail(it == s.end());
        s.erase(it);
      }     
    }
    if (s.find(-x) != s.end())
      s.erase(-x);
  }
  cout << "Possible\n";
}