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
#prag\
ma GCC optimize("Ofast")
#include <bits/stdc++.h>
typedef long long ll;
#define N 100005
using namespace std;
int n,m,a[N],fa[N],id[N],tmp[N];
bool visited[N];
int Find(int x)
{
	if(fa[x]!=x) fa[x]=Find(fa[x]);
	return fa[x]; 
}
bool cmp(int x,int y)
{
	return a[x]<a[y];
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;++i)
	{
		scanf("%d",&a[i]);
		fa[i]=i;
	}
	for(int i=1;i<=m;++i)
	{
		int u,v; scanf("%d%d",&u,&v); ++u; ++v;
		int fu=Find(u),fv=Find(v); fa[fv]=fu;
	}
	if(m==n-1)
	{
		printf("0\n");
		return 0;
	}
	for(int i=1;i<=n;++i) if(fa[i]==i) id[i]=i;
	for(int i=1;i<=n;++i) if(a[i]<a[id[Find(i)]]) id[fa[i]]=i;
	int cnt=0;
	ll ans=0;
	for(int i=1;i<=n;++i) if(fa[i]==i) visited[id[i]]=true,ans+=a[id[i]],++cnt;
	for(int i=1;i<=n;++i) tmp[i]=i;
	sort(tmp+1,tmp+n+1,cmp);
	for(int i=1;i<=n && cnt<2*(n-m-1);++i) if(!visited[tmp[i]]) ans+=a[tmp[i]],++cnt;
	if(cnt<2*(n-m-1)) printf("Impossible\n"); else printf("%lld\n",ans);
	return 0;
}