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 Rust (1.15.1) (judge verdict: Runtime Error)

#![allow(non_snake_case)]
#![allow(unused_variables)]

fn main() {
    let N: usize = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.trim().parse().unwrap()
    };
    let a: Vec<usize> = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.split_whitespace()
            .map(|x| x.parse().unwrap())
            .collect()
    };

    // 最小値を根とする木を考えればよい
    let mut c = std::collections::HashMap::new();
    for i in &a { *(c.entry(i).or_insert(0)) += 1; }
    let m = *a.iter().min().unwrap();
    let ans = if c[&m] == 1 && ((m + 1)..(2 * m + 1)).all(|i| c[&i] >= 2) && ((2 * m + 1)..N).all(|i| !c.contains_key(&i)) ||
        c[&m] == 2 && ((m + 1)..(2 * m)).all(|i| c[&i] >= 2) && ((2 * m)..N).all(|i| !c.contains_key(&i)) {
            "Possible"
        } else {
            "Impossible"
        };

    println!("{}", ans);
}