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;
int n,m;
vector<int>G[1<<17];
long a[1<<17];
int color[1<<17];
void dfs(int u,int c)
{
	color[u]=c;
	for(int i=0;i<G[u].size();i++)
	{
		if(!color[G[u][i]])dfs(G[u][i],c);
	}
}
int main()
{
	cin>>n>>m;
	for(int i=0;i<n;i++)cin>>a[i];
	for(int i=0;i<m;i++)
	{
		int x,y;cin>>x>>y;
		G[x].push_back(y);
		G[y].push_back(x);
	}
	int col=1;
	for(int i=0;i<n;i++)
	{
		if(!color[i])dfs(i,col++);
	}
	if(col==2)
	{
		cout<<0<<endl;
		return 0;
	}
	if(n<2*(n-m-1))
	{
		cout<<"Impossible"<<endl;
		return 0;
	}
	vector<vector<long> >t(col);
	vector<long>s;
	long ans=0;
	for(int i=0;i<n;i++)
	{
		t[color[i]].push_back(a[i]);
	}
	int cnt=0;
	for(int i=1;i<col;i++)
	{
		sort(t[i].begin(),t[i].end());
		ans+=t[i][0];cnt++;
		for(int j=1;j<t[i].size();j++)s.push_back(t[i][j]);
	}
	sort(s.begin(),s.end());
	for(int i=0;i<2*(n-m-1)-cnt;i++)ans+=s[i];
	cout<<ans<<endl;
}
