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<cstdio>
#include<iostream>
#include<set>
#include<vector>
#define ll long long
using namespace std;
vector<int> ga[100007];
set<pair<int,int> > s,s2;
vector<set<pair<int,int> > > v;
int a[100007];
bool vis[100007];

void dfs(int x){
	for(int i=0;i<ga[x].size();i++){
		int u=ga[x][i];
		if(!vis[u]){
			vis[u]=true;
			s.insert(make_pair(a[u],u));
			dfs(u);
		}
	}
}
int main()
{
	int n,m,i;
	cin>>n>>m;
	
	if(n-m==1)
	{
		cout<<0<<endl;
		return 0;
	}
	if((n-m-1)*2>n)
	{
		cout<<"Impossible"<<endl;
		return 0;
	}
	for(i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(i=1;i<=m;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		x++;y++;
		ga[x].push_back(y);
		ga[y].push_back(x);
	}
	
	for(i=1;i<=n;i++){
		if(!vis[i]){
			vis[i]=true;
			s.insert(make_pair(a[i],i));
			dfs(i);
			v.push_back(s);
			s.clear();
		}
	}
	
	for(i=1;i<=n;i++)
        s2.insert(make_pair(a[i],i));
	ll ans=0;
	for(i=0;i<n-m;i++)
	{
		s=v[i];
		ans+=(*s.begin()).first;
		s2.erase(*s.begin());
	}
	
	for(i=1;i<=2*(n-m-1)-(n-m);i++)
	{
		ans+=(*s2.begin()).first;
		s2.erase(*s2.begin());
	}
	cout<<ans<<endl;
	return 0;
}