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<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
long long n,m,a[100005],b[100005],fa[100005],ans=0,cnt=0;
inline long long find(long long x){
	return fa[x]?fa[x]=find(fa[x]):x;
}
inline long long max(long long x,long long y){
	return x>y?x:y;
}
inline long long read(){
	char c=getchar();
	for(;c>'9'||c<'0';) c=getchar();
	long long x=0;
	for(;c>='0'&&c<='9';){
		x=(x<<3)+(x<<1)+c-'0';
		c=getchar();
	}
	return x;
}
int main(){
	n=read(),m=read();
	if(m==n-1){
		puts("0");
		return 0;
	}
	for(long long i=1;i<=n;i++)
		a[i]=read(),ans+=a[i];
	for(long long i=1;i<=m;i++){
		long long x=read(),y=read();
		x=find(x+1),y=find(y+1);
		ans-=max(a[x],a[y]);
		b[++cnt]=max(a[x],a[y]);
		if(a[x]<a[y]) fa[y]=x;
		else fa[x]=y;
	}
	if(n-m>cnt+2){
		puts("Impossible");
		return 0;
	}
	sort(b+1,b+cnt+1);
	for(long long i=1;i<=n-m-2;i++) ans+=b[i];
	printf("%lld\n",ans);
	return 0; 
}