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)

//2017-10-19
//miaomiao
//I lack delicious yogurt
//
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
 
using namespace std;
 
#define For(i, a, b) for(int i = (a); i <= (int)(b); ++i)
#define N (100+5)

int a[N], tot[N];

int main(){
	int n;

	scanf("%d", &n);
	For(i, 1, n) scanf("%d", &a[i]);
	
	sort(a+1, a+n+1);
	if(a[n] != a[n-1] || a[1] < (a[n]+1)/2){puts("Impossible"); return 0;}
	For(i, 1, n) ++tot[a[i]];
	
	int mn = (a[n]+1) / 2;
	if(!(a[n] & 1) && tot[mn] != 1){puts("Impossible"); return 0;}
	if( (a[n] & 1) && tot[mn] != 2){puts("Impossible"); return 0;}

	For(i, mn+1, a[n]) if(tot[i] < 2){
		puts("Impossible"); return 0;
	}
	
	puts("Possible");

	return 0;
}