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<cstring>
#include<algorithm>
#define maxn 100005
#define LL long long
using namespace std;
int n,m,fa[maxn],use[maxn],cnt;
LL ans;
struct node { int w;bool vis; } a[maxn];
bool cmp(node p,node q) 
{
	if(p.vis) return false;
	if(q.vis) return true;
	return p.w<q.w; 
}
int find(int x)
{
	if(fa[x]==x) return x;
	return fa[x]=find(fa[x]);
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i].w);
		fa[i]=i;use[i]=i;
		a[i].vis=true;
	}
	for(int i=1;i<=m;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		u++,v++;
		int r1=find(u),r2=find(v);
		if(r1!=r2)
		{
			if(a[use[r2]].w<a[use[r1]].w)
			{
				a[use[r1]].vis=false;
				use[r1]=use[r2];
			}
			else a[use[r2]].vis=false;
			fa[r2]=r1;
		}
	}
	for(int i=1;i<=n;i++)
		if(find(i)==i) 
		{
			ans+=1ll*a[use[i]].w;
			cnt++;
		}
	if(cnt==1)
	{
		printf("0\n");
		return 0;
	}
	sort(a+1,a+n+1,cmp);
	for(int i=1;i<=cnt-2;i++)
		if(a[i].vis)
		{
			printf("Impossible\n");
			return 0;
		}
		else ans+=1ll*a[i].w;
	printf("%lld\n",ans);
}