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
/***********************
*  Author: xuziyuan    * 
***********************/

#include <bits/stdc++.h>

#define rt0 return 0
#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define replet(c) for(char c='a';c<='z';c++)
#define LL long long
#define pii pair <int,int>
#define pb push_back
#define fi first
#define se second
#define mpr make_pair
#define sqr(a) ((a)*(a))

using namespace std;

const LL MOD=1e9+7;

LL n,m,a[100010],par[100010],con=0,ans=0;
vector <LL> fol[100010],all;

LL FIND(LL pos)//dsu
{
	if(par[pos]!=pos) par[pos]=FIND(par[pos]);
	return par[pos];
}

inline void UNION(LL pos1,LL pos2)
{
	par[FIND(pos1)]=FIND(pos2);
}

int main()
{
	rep(i,100005) par[i]=i;
	cin>>n>>m;
	rep(i,n) scanf("%I64d",&a[i]);
	rep(i,m)
	{
		LL x,y;
		scanf("%I64d%I64d",&x,&y);
		UNION(x,y);
	}
	rep(i,n)
	{
		if(FIND(i)==i) con++;
		fol[FIND(i)].pb(a[i]);
	}
	if(con==1)
	{
		puts("0");
		rt0;
	}
	if(con*2-2>n)
	{
		puts("Impossible");
		rt0;
	}
	rep(i,n)
	{
		if(fol[i].size()==0) continue;
		sort(fol[i].begin(),fol[i].end());
		ans+=fol[i][0];
		repn(j,fol[i].size()-1) all.pb(fol[i][j]);
	}
	sort(all.begin(),all.end());
	rep(i,con-2) ans+=all[i];
	cout<<ans<<endl;
	rt0;
}