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 int long long
using namespace std;

inline int read()
{
	int sum=0,ff=1; char ch=getchar();
	while(!isdigit(ch))
	{
		if(ch=='-') ff=-1;
		ch=getchar();
	}
	while(isdigit(ch))
		sum=sum*10+(ch^48),ch=getchar();
	return sum*ff;
}

const int N=1e5+5;

int n,m,fa[N],val[N],al[N],hh[N],ans,tot,gs;
vector<int> g[N];

inline int find(int x)
{
	if(x==fa[x]) return x;
	return fa[x]=find(fa[x]);
}

signed main()
{
	n=read();
	m=read();
	for ( int i=1;i<=n;i++ ) 
	{
		val[i]=read();
		fa[i]=i;
	}
	for ( int i=1;i<=m;i++ )
	{
		int x,y;
		x=read()+1,y=read()+1;
		x=find(x),y=find(y);
		if(x==y) continue;
		fa[x]=y;
	}
	for ( int i=1;i<=n;i++ ) 
	{
		int x=find(i);
		if(!al[x]) al[x]=++tot;
		g[al[x]].push_back(val[i]);
	}	
	if(tot==1) 
		return printf("0\n"),0;
	for ( int i=1;i<=tot;i++ ) sort(g[i].begin(),g[i].end());
	for ( int i=1;i<=tot;i++ ) ans+=g[i][0];
	for ( int i=1;i<=tot;i++ )
		for ( int j=1;j<g[i].size();j++ ) hh[++gs]=g[i][j];
	if(gs<tot-2) return printf("Impossible\n"),0; 
	sort(hh+1,hh+gs+1);
	for ( int i=1;i<=tot-2;i++ ) ans+=hh[i];
	printf("%lld\n",ans);
	return 0;
}