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>
#include<boost/variant.hpp>
using namespace std;
typedef long long ll;
typedef vector<boost::variant<bool, ll, int, string, double, char*, const char*>> any;
template<typename T> inline void pr(const vector<T> &xs){
	for(int i=0; i<xs.size()-1; i++) cout<<xs[i]<<" ";
	(xs.empty()?cout:(cout<<xs[xs.size()-1]))<<endl;
}
#ifdef DEBUG
#define debug(...) pr(any{__VA_ARGS__})
#define debugv(x) pr((x))
#else
#define debug(...)
#define debugv(x)
#endif

int main(){
	int N;
	cin >> N;
	vector<int> A(N);
	for(int i=0; i<N; i++){
		cin >> A[i];
	}

	vector<int> cnt(N);
	for(int i=0; i<N; i++) cnt[A[i]]++;

	int i=1;
	while(i<N&&cnt[i]==0) i++;
	if(i==1){
		if(cnt[i]==1&&cnt[i+1]==N-1){
			cout << "Possible" << endl;
		} else if(cnt[i]==2&&N==2){
			cout << "Possible" << endl;
		} else {
			cout << "Impossible" << endl;
		}
		return 0;
	}

	if(cnt[i]>2){
		cout << "Impossible" << endl;
		return 0;
	}

	i++;
	while(i<N&&cnt[i]>=2) i++;
	while(i<N){
		if(cnt[i]!=0) {
			cout << "Impossible" << endl;
			return 0;
		}
		i++;
	}

	cout << "Possible" << endl;

	return 0;
}