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<cstdio>
#include<algorithm>
#include<vector>
#define N_ 101000
using namespace std;
vector<int>E[N_], TP;
vector<int>V;
vector<vector<int> >U;
int n, m, v[N_], w[N_];
void DFS(int a) {
	v[a] = 1;
	TP.push_back(w[a]);
	int i;
	for (i = 0; i < E[a].size(); i++)if (!v[E[a][i]])DFS(E[a][i]);
}
int main() {
	int i, a, b, j;
	scanf("%d%d", &n,&m);
	if (m == n - 1) {
		printf("0\n");
		return 0;
	}
	for (i = 1; i <= n; i++)scanf("%d", &w[i]);
	for (i = 1; i <= m; i++) {
		scanf("%d%d", &a, &b);
		a++, b++;
		E[a].push_back(b);
		E[b].push_back(a);
	}
	int S = 0, Com = 0;
	for (i = 1; i <= n; i ++) {
		if (!v[i]) {
			TP.clear();
			DFS(i);
			sort(TP.begin(), TP.end());
			Com++;
			U.push_back(TP);
		}
	}
	long long res = 0;
	for (i = 0; i < U.size(); i++) {
		res += U[i][0];
		for (j = 1; j < min((int)U[i].size(), Com - 1); j++) {
			V.push_back(U[i][j]);
		}
	}
	if (Com * 2 - 2 > (int)V.size() + Com) {
		printf("Impossible\n");
	}
	else {
		sort(V.begin(), V.end());
		for (j = 0; j < Com * 2 - 2 - Com; j++) {
			res += V[j];
		}
		printf("%lld\n", res);
	}
}