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;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef set<int> si;
typedef pair<int,int> pii;
#define rep(i,a,b) for(int i=(a); i<(b); ++i)
#define per(i,a,b) for(int i=(b)-1; i>=(a); --i)
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define F first
#define S second

int main(void) {
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	//freopen("in.txt", "r", stdin);
	int n, m;
	cin >> n >> m;
	vl a(n);
	rep(i,0,n) {
		cin >> a[i];
	}
	vector<vi> adj(n);
	rep(i,0,m) {
		int x,y;
		cin >> x >> y;
		adj[x].pb(y);
		adj[y].pb(x);
	}
	vi id(n, -1);
	int cnt = 0;
	rep(i,0,n) {
		if(id[i] != -1) continue;
		id[i] = cnt++;
		queue<int> q;
		q.push(i);
		while(!q.empty()) {
			int u = q.front();
			q.pop();
			for(auto v : adj[u]) {
				if(id[v] == -1) {
					id[v] = id[u];
					q.push(v);
				}
			}
		}
	}
	if(cnt == 1) {
		cout << 0 << '\n';
		return 0;
	}
	vector<vl> b(cnt);
	rep(i,0,n) {
		b[id[i]].pb(a[i]);
	}
	if((cnt-1)*2 > n) {
		cout << "Impossible" << '\n';
	} else {
		ll ret = 0;
		vl cand;
		rep(i,0,cnt) {
			sort(all(b[i]));
			ret += b[i][0];
			rep(j,1,sz(b[i])) {
				cand.pb(b[i][j]);
			}
		}
		sort(all(cand));
		int k = (cnt-1)*2-cnt;
		rep(i,0,k) {
			ret += cand[i];
		}
		cout << ret << '\n';
	}
	return 0;
}