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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define max(p,q)((p)>(q)?(p):(q))

int n,a[110];
int main(){
	scanf("%d",&n);
	int M=0;
	rep(i,0,n){
		int t;
		scanf("%d",&t);
		a[t]++;
		M=max(M,t);
	}
	
	int ngflag=0;
	if(M%2==0){
		rep(i,0,M/2)ngflag|=a[i]!=0;
		rep(i,M/2+1,M+1)ngflag|=a[i]<=1;
		if(a[M/2]!=1)ngflag=1;
	}else{
		if(M==1)ngflag=1;
		rep(i,0,M/2+1)ngflag|=a[i]!=0;
		rep(i,M/2+1,M+1)ngflag|=a[i]<=1;
	}
	puts(ngflag?"Impossible":"Possible");
}