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<algorithm>
#include<vector>
#define LL long long
std::vector<int> S;
std::vector<int> G[100010];
int N, M;
LL ans;
LL a[100010];
bool went[100010];
void dfs(int s, std::vector<int> &v) {
	v.push_back(a[s]);
	went[s] = true;
	for (auto u : G[s]) {
		if (!went[u])dfs(u, v);
	}
}
int main() {
	std::cin >> N >> M;
	for (int i = 0; i < N; i++) {
		std::cin >> a[i];
	}
	for (int i = 0; i < M; i++) {
		int x, y;
		std::cin >> x >> y;
		G[x].emplace_back(y);
		G[y].emplace_back(x);
	}
	if (N - 1 == M) {
		std::cout << 0 << std::endl;
		return 0;
	}
	int need = (N - M - 1) * 2;
	if (N < need) {
		std::cout << "Impossible" << std::endl;
		return 0;
	}
	for (int i = 0; i < N; i++) {
		if (!went[i]) {
			need--;
			std::vector<int> v;
			dfs(i, v);
			std::sort(v.begin(), v.end());
			ans += v[0];
			for (int i = 1; i < v.size(); i++)S.emplace_back(v[i]);
		}
	}
	std::sort(S.begin(), S.end());
	for (int i = 0; i < need; i++)ans += S[i];
	std::cout << ans << std::endl;
	return 0;
}