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 <bits/stdc++.h>
#define Int int64_t

using namespace std;

int main() {
	int N, M;
	cin >> N >> M;
	vector<Int> a(N);
	for (int i = 0; i < N; ++i) { cin >> a[i]; }
	vector<vector<int>> g(N);
	for (int i = 0; i < M; ++i) {
		int x, y;
		cin >> x >> y;
		g[x].emplace_back(y);
		g[y].emplace_back(x);
	}

	if (M == N - 1) {
		cout << "0\n";
		return 0;
	}
	if (2 * (N - M - 1) > N) {
		cout << "Impossible\n";
		return 0;
	}

	Int ans = 0;
	vector<Int> V;
	vector<bool> used(N, false);
	for (int i = 0; i < N; ++i) {
		if (used[i]) { continue; }
		
		deque<int> que;
		que.emplace_back(i);
		vector<Int> tmp;
		tmp.emplace_back(a[i]);
		while (!que.empty()) {
			int u = que.front(); que.pop_front();
			if (used[u]) { continue; }
			used[u] = true;
			
			for (auto v : g[u]) {
				if (used[v]) { continue; }
				que.emplace_back(v);
				tmp.emplace_back(a[v]);
			}
		}
		sort(tmp.begin(), tmp.end());
		ans += tmp[0];
		for (int i = 1; i < tmp.size(); ++i) {
			V.emplace_back(tmp[i]);
		}
	}

	sort(V.begin(), V.end());
	for (int i = 0; i < N - M - 2; ++i) { ans += V[i]; }
	cout << ans << endl;

	return 0;
}
