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)

///*BY ME*///
#include <bits/stdc++.h>
using namespace std;
#define y1 sijf
#define RampageRead ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);mt19937 rng(std::chrono::system_clock::now().time_since_epoch().count())
#define all(v) (v).begin(),(v).end()
#define pb push_back
#define F first
#define S second
#define endl '\n'
#define flsh cout<<flush
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef double dd;
const ll MOD=1e9+7;
set<int>ans;
int n,a[101],sum,mn,mx,minn=1000000,maxx;
int main()
{
    RampageRead;
    cin>>n;
    if(n>2) mn=1+(n-1)*2;
    else mn=2;
    for(int i=n-1;i>=n/2;i--) mx+=i*2;
    if(n&1) mx-=(n/2);
    //cout<<mn<<" "<<mx<<endl;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
        sum+=a[i];
        ans.insert(a[i]);
        minn=min(minn,a[i]);
        maxx=max(maxx,a[i]);
    }
    sort(a,a+n);
    for(int i=2;i<n;i++)
    if(a[i-1]+1<a[i] or mx<sum or mn>sum or ans.size()>n/2+1 or (a[0]==1 and ans.size()>2))
        {
            cout<<"Impossible";
            return 0;
        }
    cout<<"Possible";
    return 0;
}