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>
using namespace std;

template<typename t>inline t min_(t a,t b){return a<=b?a:b;}
template<typename t>inline t max_(t a,t b){return a>=b?a:b;}
template<typename t>inline void chkmin(t&a,t b){if(a>b)a=b;}
template<typename t>inline void chkmax(t&a,t b){if(a<b)a=b;}

typedef long long ll;
const int maxn=1e5+10;
int n,m,a[maxn],fa[maxn],need;
bool vis[maxn];
priority_queue<int,vector<int>,greater<int> > q1,q2;
ll ans;

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

int main(){
	cin>>n>>m;
	if(m==n-1){
		puts("0");
		return 0;
	}
	for(int i=0;i<n;++i){
		cin>>a[i];
		fa[i]=i;
	}
	for(int i=0,u,v;i<m;++i){
		cin>>u>>v;
		u=findfa(u);v=findfa(v);
		if(a[u]<=a[v])
			fa[v]=u;
		else
			fa[u]=v;
	}
	for(int i=0;i<n;++i)
		if(findfa(i)==i)
			q1.push(a[i]);
		else
			q2.push(a[i]);
	need=(n-m-1)*2;
	for(int i=0;i<n-m;++i)
		ans+=q1.top(),q1.pop(),--need;
	while(!q1.empty())
		q2.push(q1.top()),q1.pop();
	while(need)
		if(q2.empty()){
			puts("Impossible");
			return 0;
		}
		else
			ans+=q2.top(),q2.pop(),--need;
	printf("%lld\n",ans);
	return 0;
}
