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;
typedef long long int ll;
const ll mod = 1e9+7;

int main(){
    int n;
    cin >> n;
    vector<int> a(n);
    int one = 0;
    int mx = 0;
    int now = 0;
    bool same = true;
    for(int i = 0; i < n; i++){
        cin >> a[i];
        mx = max(mx,a[i]);
    }
    sort(a.begin(),a.end());
    int cnt = 0;
    for(int i = 0; i < n-1; i++){
        if(a[i] != a[i+1]){
            same = false;
            break;
        }
    }
    if(same){
        if(a[0]*2 == n || a[0]*2+1 == n){
            cout << "Possible" << endl;
            return 0;
        } else {
            cout << "Impossible" << endl;
            return 0;
        }
    }
    if(n == 2 && !(a[0] == 1 && a[1] == 1)){
        cout << "Impossible" << endl;
        return 0;
    }

    if(a[0]*2 == cnt || a[0]*2+1 == cnt){
            cout << "Possible" << endl;
            return 0;
        } else {
            cout << "Impossible" << endl;
            return 0;
        }
    now = a[0];
    int mn = mod;
    for(int i = 0; i < n-1; i++){
        if(a[i] == a[i+1]){
            mn = a[i];
            break;
        }
    }
    int lim = mn*2-1;
    for(int i = 0; i < n; i++){
        if(a[i] == 1) one++;
        if(now+1 == a[i]) now = a[i];
        else if(now+1 < a[i] || a[i] > lim){
            cout << "Impossible" << endl;
            return 0;
        }
    }
    if(one > 1 && mx > 1){
        cout << "Impossible" << endl;
            return 0;
    }
    cout << "Possible" << endl;
    return 0;
}