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<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>

using namespace std;

const int N=109;

int n,mv=2e9;
int a[N],tot;
int cnt[N];

inline void end()
{
	puts("Impossible");
	exit(0);
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		if(a[i]>n)end();
		cnt[a[i]]++;
		mv=min(mv,a[i]);
	}	
	
	tot=n;
	if(mv>2)end();
	tot-=cnt[mv];
	for(int i=mv+1;i<=mv*2-(!(cnt[mv]&1));i++)
	{
		if((cnt[i]<=1 && tot>0))end();
		tot-=cnt[i];
	}
	if(tot==0)
		puts("Possible");
	else
		end();
	return 0;
}