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
//writer：Sakura
//problem：Forest
//id：Petrozavodsk Contest 001--D
//from：AtCoder

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <iostream>
#include <vector>
#define N 100005
#define inf 0x3f3f3f3f
using namespace std;

typedef long long ll;

int n,m,f[N],ok[N],ok_id[N],id[N],a[N];
bool b[N]; 
ll ans=0;

int find(int x){
	return f[x]==x?x:f[x]=find(f[x]);
}

bool cmp(int x,int y){
	return a[x]<a[y];
}
int main(){
	scanf("%d %d",&n,&m);
	int res=0;
	res=(n-1-m)*2;
	if(res>n){
		puts("Impossible");
		exit(0);
	}
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		f[i]=id[i]=i;
		ok[i]=inf;
	}
	for(int i=1;i<=m;i++){
		int l,r;
		scanf("%d %d",&l,&r);
		++l,++r;
		f[find(l)]=find(r);
	}
	for(int i=1;i<=n;i++){
		int now=find(i);
		if(a[i]<ok[now]){
			ok[now]=a[i],ok_id[now]=i;
		}
	}
	int c=0; 
	for(int i=1;i<=n;i++){
		if(find(i)==i){
			++c;
			ans+=ok[i];
			b[ok_id[i]]=1;
			--res;
		}
	}
	if(c==1){
		puts("0");
		exit(0);
	}
	sort(id+1,id+n+1,cmp);
	for(int i=1,j=1;i<=res;i++){
		for(;b[id[j]];j++);
		ans+=a[id[j++]];
	}
	printf("%lld\n",ans);
	return 0;
}

