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
//左手拈着花，右手舞着剑 
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstring>
#include<map>
#include<vector>
#include<queue>
#define ll long long
#define pi pair<int,int>
#define mp(a,b) make_pair(a,b)
using namespace std;
const int N=100005;priority_queue
<pi,vector<pi>,greater<pi> >q[N];
int val[N],fa[N],rk[N],n,m,k;vector<pi>v;
void take(){
	for(int i=0;i<=n;i++)
	fa[i]=i,rk[i]=0;
} int get(int x){
	return fa[x]==x?x:fa[x]=get(fa[x]);
} void merge(int x,int y){
	x=get(x);y=get(y);
	if(x==y) return ;
	if(rk[x]<rk[y]) fa[x]=y;
	else{
		fa[y]=x;if(rk[x]==rk[y]) rk[x]++;
	}
} int main(){
	ll ans=0;scanf("%d%d",&n,&m);take();
	for(int i=0;i<n;i++) scanf("%d",&val[i]);
	for(int i=0,x,y;i<m;i++){
		scanf("%d%d",&x,&y);
		int fx=get(x),fy=get(y);
		if(fx!=fy) merge(fx,fy);
	} for(int i=0;i<n;i++)
	q[get(i)].push(mp(val[i],i));
	for(int i=0;i<n;i++)
	if(!q[i].empty()) v.push_back(q[i].top()),
	val[q[i].top().second]=2e9,
	ans+=q[i].top().first;
	if(v.size()==1) puts("0");
	else if(n<(v.size()-1)*2)
	puts("Impossible");else{
		sort(val,val+n);
		n=(v.size()-1)*2-v.size();
		for(int i=0;i<n;i++) ans+=val[i];
		cout<<ans<<endl; 
	} return 0;
}