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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.util.Arrays;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Main m = new Main();
		m.run();
	}

	Scanner sc = new Scanner(System.in);

	void run() {
		int n = sc.nextInt();
		int m = sc.nextInt();
		int a[] = new int[n];

		for (int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
		}

		int uf[] = new int[n];
		for (int i = 0; i < n; i++) {
			uf[i] = i;
		}

		for (int i = 0; i < m; i++) {
			int x = sc.nextInt();
			int y = sc.nextInt();

			uf[uf[x]] = uf[y];
		}

		Map<Integer, List<Integer>> costs = new HashMap<Integer, List<Integer>>();
		for (int i = 0; i < n; i++) {
			List<Integer> list;
			int parent = find(uf, i);
			if (costs.containsKey(parent)) {
				list = costs.get(parent);
			} else {
				list = new LinkedList<Integer>();
			}
			list.add(a[i]);
			costs.put(parent, list);
		}

		int other[] = new int[n - costs.keySet().size()];
		int oi = 0;
		int res = 0;
		for (List<Integer> list : costs.values()) {

			Integer[] arr = list.toArray(new Integer[0]);
			Arrays.sort(arr);

			for(int i = 1; i < arr.length ; i++){
				other[oi] = arr[i];
				oi ++;
			}
			
			res += arr[0];
		}
		Arrays.sort(other);
		if(costs.keySet().size() == 1){
			System.out.println(0);
			return;
		}
		if(other.length < (costs.keySet().size() -1)* 2 - costs.keySet().size()){
			System.out.println("Impossible");
			return;
		}
		for(int i =0 ; i <  (costs.keySet().size() -1)* 2 - costs.keySet().size(); i++){
			res += other[i];
		}
		System.out.println(res);
		
	}


	int find(int[] uf, int i) {
		if (i != uf[i]) {
			uf[i] = find(uf, uf[i]);
		}
		return uf[i];
	}
}