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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
int n,m,a[100009],f[100009],sum,ans;
vector<int>v[100009],other;
int getf(int x) {
	if(x!=f[x]) f[x]=getf(f[x]);
	return f[x];
}
int main() {
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++) f[i]=i;
	for(int i=0;i<n;i++) scanf("%d",a+i);
	for(int i=0;i<m;i++) {
		int x,y;
		scanf("%d%d",&x,&y);
		int fx=getf(x),fy=getf(y);
		f[fy]=fx;
	}
	for(int i=0;i<n;i++) {
		v[getf(i)].push_back(a[i]);
		if(f[i]==i) sum++;
	}
	if(sum==1) {
		printf("0\n");
		return 0;
	}
	sum=sum*2-2;
	for(int i=0;i<n;i++)
		if(!v[i].empty()) {
			sort(v[i].begin(),v[i].end());
			ans+=v[i][0];
			sum--;
			for(int j=1;j<v[i].size();j++) other.push_back(v[i][j]);
		}
	if(sum>(int)other.size()) {
		printf("Impossible\n");
		return 0;
	}
	sort(other.begin(),other.end());
	for(int i=0;i<sum;i++) ans+=other[i];
	printf("%d\n",ans);
	return 0;
}