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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define se second
#define fi first
#define pb push_back
const int N=1e5 + 5;
int par[N],v[N];
int mn[N],tak[N],cu[N];
vector<pair<int, int> > q;

int findp(int u){
	if(par[u] == u) return u;
	return par[u]=findp(par[u]);
}

int main(){

//	freopen("in.txt","r",stdin);

	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=0;i < n;i++){
		scanf("%d",&v[i]);
	}
	for(int i=0;i < n;i++)
		par[i]=i, mn[i]=2e9;
	int cmp=n,e;
	for(int i=0,a,b;i < m;i++){
		scanf("%d%d",&a,&b);
		int pa=findp(a),pb=findp(b);
		if(pa != pb) par[pa]= pb, cmp--;
	}
	if(cmp == 1) return puts("0"), 0;
	e=2 * (cmp - 1);
	for(int i=0;i < n;i++){
		mn[findp(i)]=min(mn[findp(i)],v[i]);
	}
	ll ans=0;
	for(int i=0;i < n;i++){
		if(!tak[findp(i)] && v[i] == mn[findp(i)]){
			tak[findp(i)]=1;
			ans+=v[i];
			cu[i]=1;
			e--;
		}
	}
	assert(e == cmp - 2);
	for(int i=0;i < n;i++){
		if(!cu[i]) q.pb({ v[i], findp(i) });
	}
	sort(q.begin(),q.end());
	reverse(q.begin(),q.end());
	while(q.size() && e){
		int am=q.back().fi;
		q.pop_back();
		ans+=am;
		e--;
	}
	if(!e) cout << ans << endl;
	else cout << "Impossible" << endl;

	return 0;

}
