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 <vector>
#include <iostream>
#include <algorithm>
using namespace std;
vector<int> comp;
vector<vector<int> > G;
void dfs(int pos, int pre) {
	comp.push_back(pos);
	for (int i : G[pos]) {
		if (i == pre) continue;
		dfs(i, pos);
	}
}
int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(false);
	int N, M;
	cin >> N >> M;
	G.resize(N);
	vector<int> v(N);
	for (int i = 0; i < N; ++i) cin >> v[i];
	for (int i = 0; i < M; ++i) {
		int a, b;
		cin >> a >> b;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	vector<bool> vis(N);
	vector<int> all; int C = 0;
	long long ans = 0;
	for (int i = 0; i < N; ++i) {
		if (vis[i]) continue;
		comp.clear();
		dfs(i, -1);
		int opt = 1 << 30;
		for (int j : comp) {
			vis[j] = true;
			opt = min(opt, v[j]);
		}
		ans += opt;
		++C;
		for (int j : comp) {
			if (v[j] == opt) {
				opt = -1;
			}
			else {
				all.push_back(v[j]);
			}
		}
	}
	sort(all.begin(), all.end());
	if (C == 1) cout << 0 << endl;
	else if (all.size() < C - 2) cout << "Impossible" << endl;
	else {
		for (int i = 0; i < C - 2; ++i) ans += all[i];
		cout << ans << endl;
	}
	return 0;
}