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>
#pragma GCC optimize("Ofast")

# define pb push_back
# define fr first
# define sc second
# define mk make_pair

using namespace std;

const long long linf = 1e18 + 7;
const int inf = 1e9 + 7;
const int N = 1e5 + 5;

typedef long long ll;

int n, a[N], cnt[N], cn;

int main(){
      cin >> n;

      for(int i = 1; i <= n; i ++){
            cin >> a[i];
            if(a[i] == 1) cn ++;
      }

      sort(a + 1, a + n + 1);

      for(int i = n; i > 1; i --){
            if(a[i] - 1 > a[i - 1] || (a[n] + 1) / 2 > a[i]) {
                  puts("Impossible");
                  return 0;
            }
            cnt[a[i]] ++;
      }

      cnt[a[1]] ++;

      if(cnt[a[n]] < 2 || (cn > 1 && n != 2) || (cnt[a[n]] == n && a[1] > 1)){
            puts("Impossible");
            return 0;
      }

      for(int i = (a[n] + 1) / 2; i <= a[n]; i ++){
            if(!cnt[i]){
                  puts("Impossible");
                  return 0;
            }
            if(a[n] % 2 == 0 && i > (a[n] + 1) / 2){
                  if(cnt[i] < 2){
                        puts("Impossible");
                        return 0;
                  }
            }
      }

      puts("Possible");
}