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 <queue>
using namespace std;
typedef long long ll;
int N,M,a,b;
ll A[100010];
vector<vector<int>> v(100010);
vector<vector<ll>> u(100010);
int c = 0,visited[100010] = {0};

void dfs(int n){
	visited[n] = 1;
	u[c].push_back(A[n]);
	for(int i=0;i<v[n].size();i++){
		if(visited[v[n][i]]==0) dfs(v[n][i]);
	}
}

int main(){
	cin >> N >> M;
	for(int i=0;i<N;i++) cin >> A[i];
	for(int i=0;i<M;i++){
		cin >> a >> b;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	if(M==N-1){
		cout << 0 << endl;
		return 0;
	}
	if(N<2*(N-M-1)){
		cout << "Impossible" << endl;
		return 0;
	}
	for(int i=0;i<N;i++){
		if(visited[i]==0){
			dfs(i);
			c++;
		}
	}
	ll ans = 0;
	priority_queue<ll> Q;
	for(int i=0;i<c;i++){
		sort(u[i].begin(),u[i].end());
		for(int j=0;j<u[i].size();j++){
			if(j==0) ans += u[i][j];
			else Q.push(-u[i][j]);
		}
	}
	while(c<2*(N-M-1)){
		ll t = Q.top();
		ans += -t;
		Q.pop();
		c++;
	}
	cout << ans << endl;
}