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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun main(args: Array<String>) {
  readLine()!!.toInt()
  val a = readLine()!!.split(" ").map { it.toInt() }
  val mn = a.min()!!
  val mx = a.max()!!
  val c = a.count { it == mn }
  var ans = true
  if (mn == 1) {
    if (c > 1 || mn * 2 != mx) ans = false
  } else {
    if (c == 1) {
      if (2 * mn != mx) ans = false
    } else if (c == 2) {
      if (2 * mn - 1 != mx) ans = false
    } else {
      ans = false
    }
    for (i in mn + 1..mx) {
      if (a.count { it == i } < 2) ans = false
    }
  }
  println(if (ans) "Possible" else "Impossible")
}