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 <cstdio>
#include <queue>
using namespace std;
int par[100010];
int find(int x) {
	return x == par[x] ? x : (par[x] = find(par[x]));
}
int sz[100010];
priority_queue<int> v[100010];
void unite(int x, int y) {
	x = find(x);
	y = find(y);
	if (sz[x] > sz[y]) {
		swap(x, y);
	}
	par[x] = y;
	while (v[x].size()) {
		v[y].push(v[x].top()); v[x].pop();
	}
	sz[y] = v[y].size();
}
int main() {
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i++) {
		int x;
		scanf("%d", &x);
		par[i] = i;
		sz[i] = 1;
		v[i].push(-x);
	}
	for (int i = 0; i < m; i++) {
		int u, v;
		scanf("%d%d", &u, &v);
		unite(u, v);
	}
	priority_queue<pair<int, int> > q;
	for (int i = 0; i < n; i++) {
		if (find(i) == i) {
			q.push(make_pair(v[i].top(), i));
		}
	}
	long long ans = 0;
	while (q.size() > 1) {
		pair<int, int> a, b;
		a = q.top(); q.pop();
		b = q.top(); q.pop();
		int x = a.second, y = b.second;
		if (v[x].empty() || v[y].empty()) {
			puts("Impossible");
			return 0;
		}
		v[x].pop(); v[y].pop();
		ans += -a.first - b.first;
		unite(x, y);
		int z = find(x), c = v[z].empty() ? 0 : v[z].top();
		q.push(make_pair(c, z));
	}
	printf("%lld\n", ans);
	return 0;
}