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 <bits/stdc++.h>
using namespace std;
typedef pair<int, int> pii;
int a[105];
int cnt[105];
int main(){
	int n;
	scanf("%d", &n);
	int mx = 0;
	int mn = 1e9;
	for(int i=1;i<=n;i++){
		scanf("%d", &a[i]);
		cnt[a[i]]++;
		mx = max(mx, a[i]);
		mn = min(mn, a[i]);
	}
	if((mx+1)/2 != mn) return !printf("Impossible\n");
	if(mx%2 == 0 && cnt[mn] != 1) return !printf("Impossible\n");
	if(n == 2){
		if(a[1] == a[2] && a[1] == 1) printf("Possible\n");
		else printf("Impossible\n");
		return 0;
	}
	for(int i=mx;i>=0;i--){
		if(i == mx && cnt[i] < 2) return !printf("Impossible\n");
		else if(i != mx){
			if(cnt[i] <= cnt[i+1]) continue;
			else return !printf("Impossible\n");
		}
	}

	printf("Possible\n");
}