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 N 100005
#define ll long long
using namespace std;
int n,m;
int v[N],fa[N],fd[N];
int find(int x)
{
	if(x==fa[x])return x;
	return fa[x]=find(fa[x]);
}
int p[N];
bool cmp(int x,int y)
{
	return v[x]<v[y];
}
vector<int>vt[N];
int main()
{
	cin>>n>>m;
	for(int i=1;i<=n;i++)cin>>v[i],fa[i]=i;
	int d=0;
	for(int i=1;i<=m;i++)
	{
		int t1,t2;
		cin>>t1>>t2;
		t1++;t2++;
		int a=find(t1),b=find(t2);
		if(a!=b)fa[a]=b,d++;
	} 
	if(d==n-1)return puts("0"),0;
	int nd=(n-1-d)*2;
	if(nd>n)
	{
		puts("Impossible");
	}
	else
	{
		ll ans=0;
		for(int i=1;i<=n;i++)p[i]=i;
		sort(p+1,p+n+1,cmp);
		for(int i=1;i<=n;i++)
		{
			vt[find(i)].push_back(i);
		}
		for(int i=1;i<=n;i++)
		{
			if(i==fa[i])
			{
				int mn=0x3f3f3f3f,id=0;
				for(int j=0;j<vt[i].size();j++)
				{
					if(v[vt[i][j]]<mn)
					{
						mn=v[vt[i][j]];id=vt[i][j];
					}
				}
				ans+=mn;fd[id]=1;nd--;
			}
		}
		for(int i=1;i<=n;i++)
		{
			if(nd&&!fd[p[i]])
			{
				ans+=v[p[i]];
				fd[p[i]]=1;
				nd--;
			}
		}
		cout<<ans<<endl;
	}
	return 0;
}
