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 = 100005;

multiset<int> s[N];
int cnt;
priority_queue< int, vector<int>, greater<int> > rem;
int n, m, par[N], a[N];
long long ans;

int anc(int p) { return p == par[p] ? p : par[p] = anc(par[p]); }
void join(int p, int q) {
	p = anc(p), q = anc(q);
	assert(p != q);
	if (s[p].size() > s[q].size()) swap(p, q);
	// p->q
	for (auto i : s[p]) s[q].insert(i);
	s[p].clear(); par[p] = q;
}

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin >> n >> m;
	if (n < 2 * (n - m - 1)) return cout << "Impossible\n", 0;
	for (int i = 1; i <= n; ++i) par[i] = i;
	for (int i = 1; i <= n; ++i) cin >> a[i], s[i].insert(a[i]);

	for (int i = 0; i < m; ++i) {
		int u, v; cin >> u >> v; ++u; ++v;
		join(u, v);
	}

	for (int i = 1; i <= n; ++i) if (i == anc(i)) {
		if (++cnt <= 2 * (n - m - 1)) ans += *s[i].begin();
		s[i].erase(s[i].begin());
		for (int j : s[i]) rem.push(j);
	}
	for (int i = 1; i <= 2 * (n - m - 1) - cnt; ++i) {
		ans += rem.top(); rem.pop();
	}

	cout << ans << endl;
}