Problem:
Score : 600 points

Problem Statement

You are given a forest with N vertices and M edges. The vertices are numbered 0 through N-1.
The edges are given in the format (x_i,y_i), which means that Vertex x_i and y_i are connected by an edge.

Each vertex i has a value a_i.
You want to add edges in the given forest so that the forest becomes connected.
To add an edge, you choose two different vertices i and j, then span an edge between i and j.
This operation costs a_i + a_j dollars, and afterward neither Vertex i nor j can be selected again.

Find the minimum total cost required to make the forest connected, or print Impossible if it is impossible.

Constraints

1 ≤ N ≤ 100,000

0 ≤ M ≤ N-1

1 ≤ a_i ≤ 10^9

0 ≤ x_i,y_i ≤ N-1

The given graph is a forest.

All input values are integers.

Input

Input is given from Standard Input in the following format:

N M
a_0 a_1 .. a_{N-1}
x_1 y_1
x_2 y_2
:
x_M y_M

Output

Print the minimum total cost required to make the forest connected, or print Impossible if it is impossible.

Sample Input 1

7 5
1 2 3 4 5 6 7
3 0
4 0
1 2
1 3
5 6

Sample Output 1

7

If we connect vertices 0 and 5, the graph becomes connected, for the cost of 1 + 6 = 7 dollars.

Sample Input 2

5 0
3 1 4 1 5

Sample Output 2

Impossible

We can't make the graph connected.

Sample Input 3

1 0
5

Sample Output 3

0

The graph is already connected, so we do not need to add any edges.

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

// Try AtCoder
// author: Leonardone @ NEETSDKASU

fn main() {
    let mut stdin = String::new();
    std::io::Read::read_to_string(
    	&mut std::io::stdin(),
        &mut stdin).unwrap();
	let mut stdin = stdin.split_whitespace();
    let mut get = || stdin.next().unwrap();
    macro_rules! get {
    	() => (get().parse::<i64>().unwrap());
        ($t:ty) => (get().parse::<$t>().unwrap());
    }
    
    let n = get!(usize);
    let m = get!(usize);
    let mut a = vec![0; n];
    for i in 0..n {
    	a[i] = get!();
    }
    let mut graph = vec![vec![]; n];
    for _ in 0..m {
    	let x = get!(usize);
    	let y = get!(usize);
        graph[x].push(y);
        graph[y].push(x);
    }
    let mut group = vec![0; n];
    let mut gid = 0;
    for i in 0..n {
    	if group[i] > 0 {
        	continue;
        }
        gid += 1;
        group[i] = gid;
        let mut stk = vec![i];
        while let Some(p) = stk.pop() {
        	for &e in &graph[p] {
            	if group[e] == 0 {
                	group[e] = gid;
                    stk.push(e);
                }
            }
        }
    }
    
    let mut ns = vec![vec![]; gid+1];
    for i in 0..n {
    	let g = group[i];
        ns[g].push(i);
    }
    use std::collections::BinaryHeap;
    let mut heap = BinaryHeap::new();
    let mut idx = vec![0; gid+1];
    for g in 1..gid+1 {
    	idx[g] = 1;
    	ns[g].sort();
        let p = ns[g][0];
        let c = ns[g].len() - 1;
        heap.push((n - c, c, a[p], Box::new(vec![g])));
    }
    
    let mut cost = 0;
    let mut flag = false;
    while let Some((_, c0, a0, g0)) = heap.pop() {
    	if flag {
        	println!("Impossible");
            return;
        }
    	if let Some((_, c1, a1, g1)) = heap.pop() {
			cost += a0 + a1;
            let mut g = Vec::with_capacity(g0.len()+g1.len());
            for e in g0.iter() { g.push(*e); }
            for e in g1.iter() { g.push(*e); }
            let mut p = 0;
            let mut c = 0;
            for &e in &g {
            	let i = idx[e];
                if i < ns[e].len() {
                	let k = ns[e][i];
                	if p == 0 || a[k] < c {
                    	p = e;
                        c = a[k];
                    }
                }
            }
            if p > 0 {
            	idx[p] += 1;
                let c3 = c0+c1+3;
                heap.push((n-c3, c3, c, Box::new(g)));
            } else {
            	flag = true;
            }
        }
    }
    
    println!("{}", cost);
}