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 ll long long
using namespace std;
inline int rd(){
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-') f=0;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}
	return f?x:-x;
}
int n,m,fa[100050],vis[100050],cnt,p[100050],ct;
long long val[100050],ans;
inline int find(int x){
	if(x!=fa[x]) return fa[x]=find(fa[x]);
	return fa[x];
}
inline void unionn(int x,int y){
	x=find(x);y=find(y);fa[x]=y;
}
vector<int>G[100050];
int main(){
	n=rd(),m=rd();
	for(int i=1;i<=n;i++){
		val[i]=rd(),fa[i]=i;
	}int x,y;
	for(int i=1;i<=m;i++){
		x=rd()+1,y=rd()+1;
		unionn(x,y);
	}
	if(m==n-1){
		puts("0");
		return 0;
	}
	if((n-m-1)*2>n){
		puts("Impossible");
		return 0;
	}
	for(int i=1;i<=n;i++){
		if(vis[find(i)]) G[vis[fa[i]]].push_back(val[i]);
		else G[vis[find(i)]=++cnt].push_back(val[i]);
	}
	for(int i=1;i<=cnt;i++){
		sort(G[i].begin(),G[i].end());
		ans+=G[i][0];int sz=G[i].size();
		for(int j=1;j<sz;j++){
			p[++ct]=G[i][j];
		} 
	}sort(p+1,p+1+ct);
	for(int i=1;i<=n-m-2;i++){
		ans+=p[i];                  
	}
	cout<<ans;
	return 0;
}