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: Accepted)

#include <bits/stdc++.h>

using namespace std;

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

int main()
{
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int n;
    cin >> n;
    vector<int> a(n);
    map<int, int> cnt;
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
        ++cnt[a[i]];
    }


    sort(a.rbegin(), a.rend());

    int minvalue;
    int maxvalue;
    //build diametr
    maxvalue = a[0];
    minvalue = (maxvalue + 1) / 2;
    for (int i = maxvalue; i > minvalue; --i) {
        cnt[i] -= 2;
        if (cnt[i] < 0)
            imp();
    }

    if (maxvalue % 2 == 0)
        cnt[minvalue] -= 1;
    else
        cnt[minvalue] -= 2;

    if (cnt[minvalue] < 0)
        imp();

    // check other

    for (int i = 0; i < n; ++i) {
        if (a[i] <= minvalue && cnt[a[i]] > 0)
            imp();
    }

    cout << "Possible" << endl;


    return 0;
}