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;

#define pb push_back
#define mp make_pair
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef complex<double> point;
#define xx real()
#define yy imag()

#define REP(i, a, b) for(int i = (a); i < (int)(b); i++)
#define REPN(i, a, b) for(int i = (a); i <= (int)(b); i++)
#define FA(it, x) for(__typeof((x).begin()) it = (x).begin(); it != (x).end(); it++)
#define SZ(x) (int)(x).size()
#define BE(x) (x).begin(), (x).end()
#define SORT(x) sort(BE(x))
#define _1 first
#define _2 second

#define x1 gray_cat_x1
#define y1 gray_cat_y1

template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }

#define file "cycle"

const double EPS = 1e-9;
const double PI = acos(-1.);
const int INF = 1e9;
const ll MOD = 1e9 + 7;

const int MAXN = 1e5 + 5;

int cnt[MAXN];

void solve(){
	int n, a, mx, mn;
	scanf("%d", &n);
	mx = 0;
	mn = INF;
	REP(i, 0, n){
		scanf("%d", &a);
		cnt[a]++;
		mx = max(mx, a);
		mn = min(mn, a);
	}
	if (cnt[mx] == 1){
		printf("Impossible\n");
		return;
	}
	if (mn * 2 < mx || mn * 2 > mx + 1){
		printf("Impossible\n");
		return;
	}
	if (mn * 2 == mx && (cnt[mn] != 1 || cnt[mn + 1] == 1)){
		printf("Impossible\n");
		return;
	}
	if (mn * 2 == mx + 1 && cnt[mn] != 2){
		printf("Impossible\n");
		return;
	}
	for(int i = mx; i > mn; i--){
		if (cnt[i] < cnt[i - 1]){
			printf("Impossible\n");
			return;
		}
	}
	printf("Possible\n");
}	

int main(){

	//freopen(file".in", "r", stdin); freopen(file".out", "w", stdout);
	int t = 1;
	//cin >> t;
	while(t--){
		solve();	
	}
}