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.

Examples

Input

7 5
1 2 3 4 5 6 7
3 0
4 0
1 2
1 3
5 6


Output

7


Input

5 0
3 1 4 1 5


Output

Impossible


Input

1 0
5


Output

0
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

const int N = 100 * 1000 + 5;
int nparts = 0;

int a[N];
vector <int> adj[N];
int part[N];
vector <int> vec[N];
vector <int> rm;

void dfs(int v) {
	for (auto u : adj[v]) {
		if (!part[u]) {
			part[u] = part[v];
			dfs(u);
		}
	}
}
int main() {
	long long ans = 0;
	int n, m;
	cin >> n >> m;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	for (int i = 0; i < m; i++) {
		int u, v;
		cin >> u >> v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	if (m == n - 1) {
		cout << 0 << endl;
		return 0;
	}
	for (int i = 0; i < n; i++) {
		if (!part[i]) {
			nparts++;
			part[i] = nparts;
			dfs(i);
		}
	}
	for (int i = 0; i < n; i++) {
		vec[part[i]].push_back(a[i]);
	}
	for (int i = 1; i <= nparts; i++) {
		sort(vec[i].begin(), vec[i].end());
		ans += vec[i][0];
		for (int j = 1; j < vec[i].size(); j++) {
			rm.push_back(vec[i][j]);
		}
	}
	sort(rm.begin(), rm.end());
//	cout << nparts << endl;
	if (2 * (nparts - 1) > n) {
		cout << "Impossible" << endl;
	}
	else {
		for (int i = 0; i < nparts - 2; i++) {
			ans += rm[i];
		}
		cout << ans << endl;
	}
	return 0;
}