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

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using pii = pair<int, int>;
using ld = long double;

template <typename T>
int sz(T &x) {
  return (int)(x.size());
}

const int inf = 1'000'000'007;

void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());

  vector<int> has(n);
  for (int i = 0; i < n; ++i) {
    has[a[i]] = 1;
  }
  int ok = a[n - 1] == a[n - 2];
  ok &= a[0] >= (a[n - 1] + 1) / 2;
  for (int i = (a[n - 1] + 1) / 2; i <= a[n - 1]; ++i) {
    ok &= has[i];
  }

  if (ok) {
    cout << "Possible\n";
  } else {
    cout << "Impossible\n";
  }
}

int main() {
  ios_base :: sync_with_stdio(false);
  cin.tie(nullptr);

  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}