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
#prag\
ma GCC optimize("O3")
#include<bits/stdc++.h>
#define ll long long
using namespace std;

vector<set<pair<int,int> > > g;
vector<int> gv[100001];
set<pair<int,int> > s,s2;
bool vis[100001];
int a[100001];

inline void dfs(int x){
	for(int i=0;i<gv[x].size();i++){
		if(!vis[gv[x][i]]){
			vis[gv[x][i]]=1;
			s.insert(make_pair(a[gv[x][i]],gv[x][i]));
			dfs(gv[x][i]);
		}
	}
}

int main()
{
	ios::sync_with_stdio(false);
	int n,m;
	
	cin>>n>>m;
	
	if(n-1==m){
		cout<<0;
		return 0;
	}
	
	if((n-1-m)*2>n){
		cout<<"Impossible";
		return 0;
	}
	
	for(int i=1;i<=n;i++) cin>>a[i];
	
	for(int i=1;i<=m;i++){
		int u,v;
		cin>>u>>v;
		++u,++v;
		gv[u].push_back(v);
		gv[v].push_back(u);
	}
	
	for(int i=1;i<=n;i++)
		if(!vis[i]){
			vis[i]=1;
			s.insert(make_pair(a[i],i));
			dfs(i);
			g.push_back(s);
			s.clear();
		}
		
	for(int i=1;i<=n;i++) s2.insert(make_pair(a[i],i));
	
	ll ans=0;
	for(int i=0;i<n-m;i++){
		s=g[i];
		ans+=(ll)(*s.begin()).first;
		s2.erase(*s.begin());
	}
	
	for(int i=1;i<=2*(n-m-1)-(n-m);i++){
		ans+=(ll)(*s2.begin()).first;
		s2.erase(s2.begin());
	}
	
	cout<<ans;
}