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;
inline int read(){
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x*f;
}
#define MN 100005
int n,m,fa[MN],vis[MN],cnt,p[MN],ct;
long long a[MN],ans;
int findf(int x){return fa[x]==x?fa[x]:fa[x]=findf(fa[x]);}
void union_(int x,int y){x=findf(x);y=findf(y);fa[x]=y;return;}
vector<int> pic[MN];
int main(){
	n=read(),m=read();
	register int i,j,x,y;
	for(i=1;i<=n;i++) a[i]=read(),fa[i]=i;
	for(i=1;i<=m;i++){
		x=read()+1,y=read()+1;
		union_(x,y);
	}
	if(m==n-1) return puts("0"),0;
	if((n-m-1)*2>n) return puts("Impossible\n"),0;
	for(i=1;i<=n;i++){
		if(vis[findf(i)]) pic[vis[fa[i]]].push_back(a[i]);
		else pic[vis[findf(i)]=++cnt].push_back(a[i]);
	}
	for(i=1;i<=cnt;i++){
		sort(pic[i].begin(),pic[i].end()),ans+=pic[i][0];
		for(j=1;j<pic[i].size();j++) p[++ct]=pic[i][j];
	}
	sort(p+1,p+ct+1);
	for(i=1;i<=n-m-2;i++) ans+=p[i];
	printf("%lld",ans);
	return 0;
}