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: Accepted)

import java.util.Scanner;
import java.util.PriorityQueue;
import java.util.Arrays;
public class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		int m=sc.nextInt();
		Unionfind uf=new Unionfind(n);
		PriorityQueue<Long> q=new PriorityQueue<>();
		long[] a=new long[n];
		for(int i=0; i<n; i++){
			a[i]=sc.nextLong();
		}
		for(int i=0; i<m; i++){
			int x=sc.nextInt();
			int y=sc.nextInt();
			uf.unite(x,y);
		}
		long[] min=new long[n];
		int[] index=new int[n];
		Arrays.fill(min,1000000001);
		for(int i=0; i<n; i++){
			int root=uf.root(i);
			if(min[root]>a[i]){
				min[root]=a[i];
				index[root]=i+1;
			}
		}
		boolean[] o=new boolean[n];
		for(int i=0; i<n; i++){
			if(index[i]!=0){
				o[index[i]-1]=true;
			}
		}
		long ans=0;
		int count=0;
		for(int i=0; i<n; i++){
			if(min[i]!=1000000001){
				count++;
				ans+=min[i];
			}
			if(!o[i]){
				q.add(a[i]);
			}
		}
		if(count==1){
			System.out.println(0);
			return;
		}else if(count+count-2>n){
			System.out.println("Impossible");
			return;
		}
		for(int i=0; i<count-2; i++){
			ans+=q.poll();
		}
		System.out.println(ans);
	}
	static class Unionfind{
		int[] parent;
		public Unionfind(int n){
			this.parent=new int[n];
			for(int i = 0; i<n; i++){
				parent[i]=i;
			}
		}
		public int root(int t){
			if(parent[t]==t){
				return t;
			}else{
				return parent[t]=root(parent[t]);
			}
		}
		public void unite(int x,int y){
			if(same(x,y)){
				return;
			}
			int xroot = root(x);
			int yroot = root(y);
			parent[yroot]=xroot;
		}
		public boolean same(int x,int y){
			return root(x)==root(y);
		}
	}
}