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<cmath>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long LL;
int n,m;
LL a[100000+10];
int par[100000+10];
priority_queue<LL> mini[100000+10];

int fpar(int x){
	if (par[x]==-1)return x;
	par[x]=fpar(par[x]);
	return par[x];
}
priority_queue <LL > pq1;
LL ans=0;
int main(){
	scanf("%d%d",&n,&m);
	for (int f1=0;f1<n;f1++)scanf("%lld",&a[f1]);
	for (int f1=0;f1<n;f1++){par[f1]=-1;}
	for (int f1=0;f1<m;f1++){
		int i,j;
		scanf("%d%d",&i,&j);
		i=fpar(i);
		j=fpar(j);
		par[j]=i;
	}
	for (int f1=0;f1<n;f1++){
		int j=fpar(f1);
		mini[j].push(-a[f1]);
	}
	if (n==1+m){
		printf("0\n");
		return 0;
	}
	for (int f1=0;f1<n;f1++){
		if (par[f1]==-1){
			ans-=mini[f1].top();
			mini[f1].pop();
			while (!mini[f1].empty()){
				pq1.push(mini[f1].top());
				mini[f1].pop();
			}	
		}
	}
	
	if (int(pq1.size())>=n-m-2){
		for (int f1=1;f1<=n-m-2;f1++){
			ans-=pq1.top();
			pq1.pop();
		}
		printf("%lld\n",ans);
	} else printf("Impossible\n");
}
