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 main() {
	long long N, M, ANS = 0;
	vector<long long> A;
	vector<vector<long long> > V;
	vector<bool> B;
	cin >> N >> M;
	A.resize(N), V.resize(N), B.assign(N, false);
	for (int i = 0; i < N; i++) cin >> A[i];
	for (int i = 0; i < M; i++) {
		long long X, Y;
		cin >> X >> Y;
		V[X].push_back(Y);
		V[Y].push_back(X);
	}
	if (N > M * 2 + 2) {
		cout << "Impossible" << endl;
		return 0;
	}
	if (N - M == 1) {
		cout << 0 << endl;
		return 0;
	}
	vector<vector<long long> > F(N - M);
	long long Cur = 0;
	for (int i = 0; i < N; i++) {
		if (!B[i]) {
			stack<long long> ST;
			ST.push(i);
			while (!ST.empty()) {
				long long P = ST.top();
				ST.pop();
				B[P] = true;
				F[Cur].push_back(A[P]);
				for (int NP : V[P]) {
					if (!B[NP]) ST.push(NP);
				}
			}
			Cur++;
		}
	}
	vector<long long> F2;
	for (int i = 0; i < N - M; i++) {
		sort(F[i].begin(), F[i].end());
		ANS += F[i][0];
		for (int j = 1; j < F[i].size(); j++) F2.push_back(F[i][j]);
	}
	sort(F2.begin(), F2.end());
	for (int i = 0; i < N - M - 2; i++) {
		ANS += F2[i];
	}
	cout << ANS << endl;
}