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>
#define LL long long 
using namespace std;

const LL maxn=1e5+5;

LL ans,tot,n,m,a[maxn],b[maxn],fa[maxn],xo[maxn];
LL nw,u,v,w1,w2;
bool vis[maxn];

LL getfather(LL k)
{
	if(fa[k]==k)return k;
	return fa[k]=getfather(fa[k]);
}

int main()
{
	scanf("%lld%lld",&n,&m);
	for(int i=0;i<n;i++)
	{
		scanf("%lld",&a[i]);
		fa[i]=i;
		xo[i]=a[i];		
	}
	for(int i=0;i<m;i++)
	{
		scanf("%lld%lld",&u,&v);
		w1=getfather(u);
		w2=getfather(v);
		if(w1==w2)continue;
		fa[w1]=w2;
		xo[w1]=min(xo[w1],xo[w2]);
		xo[w2]=min(xo[w1],xo[w2]);
	}
	for(int i=0;i<n;i++)if(fa[i]==i)
	{
		b[tot]=xo[i];
		tot++;
		ans+=xo[i];
	}
	//cout<<tot<<endl;
	if(tot*2-2>n)
	{
		printf("Impossible\n");
		return 0;
	}
	if(tot==1)
	{
		printf("0\n");
		return 0;
	}
	if(tot==2)
	{
		printf("%lld\n",ans);
		return 0;
	}
	sort(b,b+tot);
	sort(a,a+n);
	nw=0;
	for(int i=0;i<n;i++)
	{
		if(a[i]<b[nw])continue;
		vis[i]=true;nw++;
		if(nw==tot)break;
	}
	nw=0;
	for(int i=0;i<n;i++)if(!vis[i])
	{
		ans+=a[i];
		nw++;
		if(nw==tot-2)break;
	}
	printf("%lld\n",ans);
	return 0;
}