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;
struct UnionFind {
	vector<int> data;
	UnionFind(int n) : data(n, -1) {}
	int root(int x) {
		return data[x] < 0 ? x : data[x] = root(data[x]);
	}
	bool merge(int x, int y) {
		x = root(x);
		y = root(y);
		if (x == y) return false;
		if (data[y] < data[x]) swap(x, y);
		data[x] += data[y];
		data[y] = x;
		return true;
	}
	int size(int x) {
		return -data[root(x)];
	}
};
int main() {
	int N, M;
	cin >> N >> M;
	vector<int> A(N);
	for (auto& x : A) cin >> x;
	UnionFind uf(N);
	for (int i = 0; i < M; ++ i) {
		int u, v;
		cin >> u >> v;
		uf.merge(u, v);
	}

	if (uf.size(0) == N) {
		cout << 0 << endl;
		return 0;
	}

	vector<priority_queue<int>> V(N);
	for (int i = 0; i < N; ++ i) {
		V[uf.root(i)].push(-A[i]);
	}
	long long r = 0;
	priority_queue<int> Q;
	int n = 0;
	for (int i = 0; i < N; ++ i) if (!V[i].empty()) {
		auto& q = V[i];
		r += -q.top(); q.pop();
		++ n;
		while (!q.empty()) {
			Q.push(q.top()); q.pop();
		}
	}
	if ((int)Q.size() < n-2) {
		cout << "Impossible" << endl;
	} else {
		for (int i = 0; i < n-2; ++ i) {
			r += -Q.top(); Q.pop();
		}
		cout << r << endl;
	}
}
