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>

#define SZ(x) ((int(x.size())))

using namespace std;

const int maxn = 110;
int M, n, t[maxn], x;

int main()
{
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		cin >> x;
		t[x]++;
		if (x > M)
			M = x;
	}
	if (M % 2 == 0)
	{
		if (t[M / 2] == 0)
		{
			cout << "Impossible";
			return 0;
		}
		else
			t[M / 2]--;
	}
	for (int i = M / 2 + 1; i <= M; i++)
	{
		if (t[i] < 2)
		{
			cout << "Impossible";
			return 0;
		}
		else
			t[i] -= 2;
	}
	for (int i = 1; i <= (M + 1) / 2; i++)
	{
		if (t[i] > 0)
		{
			cout << "Impossible";
			return 0;
		}
	}
	cout << "Possible";
	return 0;
}