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 <iostream>
#include <vector>
#include <algorithm>
#include <utility>
#define MAX_V 100010

using namespace std;
vector<long long> G[MAX_V],a;
bool used1[MAX_V],used2[MAX_V];
int V,E,k=0;
long long b[MAX_V];
pair<int,long long> p[MAX_V];

void dfs(int s){
	if(a[b[k]]>a[s]){
		b[k] = s;
	}
	used1[s] = true;
	int i = 0;
	while(true){
		int e = G[s][i];
		if(e==-1){
			break;
		}
		i++;
		if(!used1[e]){
			dfs(e);
		}
	}
}

int main(){
	int i,j;
	long long x,ans=0;
	cin >> V >> E;
	for(i=0;i<V;i++){
		cin >> x;
		a.push_back(x);
		p[i] = make_pair(x,i);
		used1[i] = false;
		used2[i] = false;
		b[i] = V;
	}
	a[V] = 1000000000;
	for(i=0;i<E;i++){
		int s,t;
		cin >> s >> t;
		G[s].push_back(t);
		G[t].push_back(s);
	}
	for(i=0;i<V;i++){
		G[i].push_back(-1);
	}
	
	for(i=0;i<V;i++){
		if(!used1[i]){
			dfs(i);
			used2[b[k]] = true;
			ans += a[b[k]];
			k++;
		}
	}
	if(k==1){
		cout << 0 << endl;
		return 0;
	}
	int cnt = k;
	if(V<2*(k-1)){
		cout << "Impossible" << endl;
		return 0;
	}
	
	sort(p,p+V);
	for(i=0;i<V;i++){
		if(cnt==2*(k-1)){
			break;
		}
		if(!used2[p[i].second]){
			ans += p[i].first;
			cnt++;
		}
	}
	cout  << ans << endl;
}
	
