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;
typedef long long LL;
const char *Yes = "Possible";
const char *No = "Impossible";
multiset < int, greater < int > > g;
multiset < int, greater < int > > :: iterator it;

int main(){
	int n, a;
	scanf("%d", &n);
	for(int i = 1; i <= n; i++){
		scanf("%d", &a);
		g.insert(a);
	}
	int len = *g.begin(); 
	for(int i = len / 2 + 1; i <= len; i++){
		for(int j = 0; j < 2; j++){
			it = g.find(i);
			if(it == g.end()){
				return puts(No), 0;
			}
			g.erase(it);
		}
	}
	if(!(len & 1)){
		it = g.find(len / 2);
		if(it == g.end()){
			return puts(No), 0;
		}
		g.erase(it);
	}
	len = (len + 1) / 2 + 1;
	for(it = g.begin(); it != g.end(); ++it){
		if(*it < len){
			return puts(No), 0;
		}
	}
	puts(Yes);
	return 0;
}