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>
#include<stdlib.h>
#include<string.h>
#include<math.h>

#define INF 1000000000
 
#define min(a,b) (((a)<(b))?(a):(b))
#define max(a,b) (((a)>(b))?(a):(b))

int compare(const void *a, const void *b) {
	return *(int*)a - *(int*)b;
}

int main() {
	int N;
	int ans = 0;
	int a[100];
	int i, j;
	
	scanf("%d", &N);
	for(i=0;i<N;i++) {
		scanf("%d", &a[i]);
	}
	
	qsort(a, N, sizeof(int), compare);
	
	for(i=1;i<N;i++) {
		if(a[i] - a[i-1] >= 2) {
			printf("Impossible\n");
			return 0;
		}
	}
	
	for(i=1;i<N-1;i++) {
		if(a[i] != a[i-1] && a[i] != a[i+1]) {
			printf("Impossible\n");
			return 0;
		}
	}
	
	if(a[0] < (a[N-1]+1)/2) {
		printf("Impossible\n");
		return 0;
	}
	
	if(N - 1 < a[N-1]) {
		printf("Impossible\n");
		return 0;
	}
	
	if(a[0] == a[2]) {
		printf("Impossible\n");
		return 0;
	}
	
	if(N > 2 && a[0] == 1 && a[1] == 1) {
		printf("Impossible\n");
		return 0;
	}
	
	if(a[N-1] != a[N-2]) {
		printf("Impossible\n");
		return 0;
	}
	
	printf("Possible\n");
	
	return 0;
}