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: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define MOD 1000000007
#define ll long long int
#define vi vector<int>
#define vii vector< vector<int> >
#define PI 3.1415926535897932384626433832795
#define INF 9223372036854775807LL

int arr[100005];

int main() {
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	int n,m;
	cin >> n >> m;
	for(int i = 0; i < n; i++) {
		cin >> arr[i];
	}
	vii graph(n+2,vi());
	for(int i = 0; i < m; i++) {
		int x,y;
		cin >> x >> y;
		graph[x].push_back(y);
		graph[y].push_back(x);
	}
	int next = 0;
	vector<multiset<int> > trees;
	int ind = 0;
	vector<bool> was (n+2,false);
	int sizsum = 0;
	while(next != n) {
		while(next < n && was[next]) {
			next++;
		}
		if(next == n) {
			break;
		}
		trees.push_back(multiset<int>());
		stack<int> st;
		st.push(next);
		was[next] = true;
		while(!st.empty()) {
			int v = st.top();
			st.pop();
			trees[ind].insert(arr[v]);
			for(int u : graph[v]) {
				if(!was[u]) {
					was[u] = true;
					st.push(u);
				}
			}
		}
		ind++;
	}
	if(n < 2*ind-2) {
		cout << "Impossible";
		return 0;
	}
	multiset<int> curr;
	multiset<pair<int,int> > ot;
	srand(time(0));
	int rn = rand()%ind;
	rn = 0;
	for(int el : trees[rn]) {
		curr.insert(el);
	}
	for(int i = 0; i < ind; i++) {
		if(i == rn) {
			continue;
		}
		for(int el : trees[i]) {
			ot.insert({el,i});
		}
	}
	ll ans = 0;
	while(ot.size() > 0) {
		int best = (*curr.begin());
		curr.erase(curr.begin());
		auto bestot = ot.begin();
		pair<int,int> pr = *bestot;
		for(int el : trees[pr.second]) {
			curr.insert(el);
			ot.erase(ot.find({el,pr.second}));
		}
		curr.erase(curr.find(pr.first));
		while(curr.empty() && !ot.empty()) {
			curr.insert(pr.first);
			for(int el : trees[pr.second]) {
				curr.erase(curr.find(el));
				ot.insert({el,pr.second});
			}
			while(pr.second == (*bestot).second) {
				bestot++;
			}
			pr = *bestot;
			for(int el : trees[pr.second]) {
				curr.insert(el);
				ot.erase(ot.find({el,pr.second}));
			}
			curr.erase(curr.find(pr.first));
		}
		ans+= best+pr.first;
	}
	cout << ans;
	return 0;
}