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;

const int N = 1e5 + 7;

long long a[N];
vector <int> ad[N];
int n, m;
int cnt;
bool mrk[N];
set < pair<long long, int> > s[N];
set < pair<long long, int> > can;
void dfs(int v) {
	mrk[v] = 1;
	s[cnt].insert({a[v], v});
	for (auto u: ad[v]) {
		if (!mrk[u])
			dfs(u);
	}
}

int main() {	
	cin >> n >> m;
	for (int i = 1; i <= n; i++)
		cin >> a[i];	
	for (int i = 1; i <= m; i++) {
		int x, y;
		cin >> x >> y;
		x++, y++;
		ad[x].push_back(y);
		ad[y].push_back(x);
	}
	for (int i = 1; i <= n; i++) {
		if (!mrk[i]) {
			cnt++;
			dfs(i);
		}
	}
	if (cnt == 1) {
		cout << 0;
		return 0;
	}
	long long res = 0;
	for (int i = 1; i <= cnt; i++) {
		res += s[i].begin()->first;
		s[i].erase(s[i].begin());
		while(s[i].size()) {
			can.insert({s[i].begin()->first, s[i].begin()->second});
			s[i].erase(s[i].begin());
		}
	}
	int tt = 2 * (cnt - 1) - cnt;
	if ((tt + cnt) % 2)
		tt++;
	if (can.size() < tt) {
		cout << "Impossible";
		return 0;
	}	
	for (int i = 1; i <= tt; i++) {
		res += can.begin()->first;
		can.erase(can.begin());
	}
	cout << res;
	}