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 <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N=100005;
int Head[N],Next[N<<1],Adj[N<<1],tot=0;
int mn[N],pos[N],a[N];
bool vis[N];
inline int read(){
	int x=0;char e=getchar();
	for (;e<'0'||e>'9';e=getchar());
	for (;e>='0'&&e<='9';e=getchar()) x=(x<<1)+(x<<3)+(e^48);
	return x;
}
inline void addedge(int u,int v){
	Next[++tot]=Head[u];
	Head[u]=tot;
	Adj[tot]=v;
	Next[++tot]=Head[v];
	Head[v]=tot;
	Adj[tot]=u;
}
inline void dfs(int x){
	mn[x]=a[x],pos[x]=x;
	vis[x]=true;
	for (int e=Head[x];e;e=Next[e])
		if (!vis[Adj[e]]){
			dfs(Adj[e]);
			if (mn[Adj[e]]<mn[x]) pos[x]=pos[Adj[e]],mn[x]=mn[Adj[e]];
		}
}
int main (){
	int n=read(),m=read();
	memset (mn,0x3f,sizeof(mn));
	for (int i=1;i<=n;i++) a[i]=read();
	for (int i=1;i<=m;i++){
		int u=read()+1,v=read()+1;
		addedge(u,v);
	}
	long long ans=0;int cnt=0;
	for (int i=1;i<=n;i++)
		if (!vis[i]){
			dfs(i);
			cnt++;
			ans+=mn[i],a[pos[i]]=1<<30;
		}
	sort(a+1,a+n+1);
	if ((cnt-1)*2>n){
		printf("Impossible");
		return 0;
	}
	for (int i=1;i<=(cnt-1)*2-cnt;i++) ans+=a[i];
	printf ("%lld",cnt==1?0:ans);
	return 0;
}