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>

using namespace std;

int const N = 1234567;

int p[N], a[N], sz[N], was[N], chosen[N];
pair<int, int> b[N];

int get(int v) {
	return v == p[v] ? v : (p[v] = get(p[v]));
}

int main() {
	ios::sync_with_stdio(false);
	int n, m;
	cin >> n >> m;
	for (int i = 0; i < n; i++) cin >> a[i];
	for (int i = 0; i < n; i++) p[i] = i;
	for (int i = 0; i < m; i++) {
		int v, u;
		cin >> v >> u;
		p[get(v)] = get(u);
	}
	for (int i = 0; i < n; i++) {
		b[i] = { a[i], i };
	}
	sort(b, b + n);
	for (int i = 0; i < n; i++) {
		sz[get(i)]++;
	}
	if (sz[get(0)] == n) {
		puts("0");
		return 0;
	}
	long long ans = 0;
	int have = 0;
	int comps = 0;
	for (int i = 0; i < n; i++) {
		int v = get(b[i].second);
		if (!was[v]) {
			was[v] = true;
			chosen[i] = true;
			ans += b[i].first;
			have++;
			comps++;
			// if (sz[v] == 1) {
			// 	++one;
			// }
		}
	}
	if (2 * (comps - 1) > n) {
		cout << "Impossible\n";
		return 0;
	}
	for (int i = 0; have < 2 * (comps - 1) && i < n; i++) {
		if (!chosen[i]) {
			ans += b[i].first;
			have++;
		}
	}
	cout << ans << endl;
}
