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<vector>
#include<algorithm>
using std::vector;
using std::sort;
long long rnum[400500];
vector <long long> G[400500];
long long f[400500],col[400500];
long long find(long long x){
	if(f[x]==x) return x;
	return f[x]=find(f[x]);
}
long long num[400500];
long long n,m,a,b;
int main(){
	scanf("%lld%lld",&n,&m);
	int cnt=0;
	for(long long i=0;i<=n+5;i++) f[i]=i;
	for(int i=1;i<=n;i++) scanf("%lld",&num[i]);
	for(int i=1;i<=m;i++){
		scanf("%lld%lld",&a,&b);
		a++; b++;
		a=find(a);b=find(b);
		if(a!=b){f[b]=a;++cnt;}
	}
	m=n-cnt;
	if(n<2*(m-1)){
		puts("Impossible");
		return 0;
	}
	cnt=0;
	if(m==1){puts("0");return 0;}
	for(int i=1;i<=n;i++){
		G[find(i)].push_back(num[i]);
		if(i==f[i]) col[++cnt]=i;
	}
	for(int i=1;i<=m;i++) sort(G[col[i]].begin(),G[col[i]].end());
	long long sum=0ll;
	long long siz=0;
	for(int i=1;i<=m;i++){
		sum+=G[col[i]][0];
		for(int j=1;j<G[col[i]].size();j++)
			rnum[++siz]=G[col[i]][j];
	}
	sort(rnum+1,rnum+siz+1);
	for(int i=1;i<=m-2;i++) sum+=rnum[i];
	printf("%lld\n",sum);
	return 0;
}