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;
using ll = long long;
vector<ll> v[100000];
bool used[100000] = {};
const ll INF = 1145148931919810;
map<ll, ll> a;
pair<ll, ll> dfs(ll now){
	used[now] = 1;
	pair<ll, ll> ret = pair<ll, ll>(a[now], now);
	for(ll next : v[now]){
		if(used[next]) continue;
		pair<ll, ll> tmp = dfs(next);
		if(tmp.first < ret.first) ret = tmp;
	}
	return ret;
}
int main(){
	ll n, m; cin >> n >> m;
	for(ll i = 0; i < n; i++){
		ll po; cin >> po;
		a[i] = po;
	}
	for(ll i = 0; i < m; i++){
		ll x, y; cin >> x >> y;
		v[x].push_back(y);
		v[y].push_back(x);
	}
	ll ans = 0;
	ll ren = 0;
	for(ll i = 0; i < n; i++){
		if(used[i]) continue;
		pair<ll, ll> tmp = dfs(i);
		ren++;
		ans += tmp.first;
		a.erase(tmp.second);
	}
	if(ren == 1){
		cout << 0 << endl;
		return 0;
	}
	if(2 * ren - 2 > n){
		cout << "Impossible\n";
		return 0;
	}
	vector<ll> po;
	for(auto ite = a.begin(); ite != a.end(); ite++) po.push_back(ite->second);
	sort(po.begin(), po.end());
	for(ll i = 0; i + ren < 2 * ren - 2; i++) ans += po[i];
	cout << ans << endl;
	return 0;
}