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<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdint>
#include<iostream>
#include<algorithm>
using namespace std;
int fa[100001];
inline int find(int x)
{
	if(fa[x]!=x)
		fa[x]=find(fa[x]);
	return fa[x];
}
bool v[100001];
struct num
{
	int x,p;
	bool operator <(num y) const
	{
		return x<y.x;
	}
}a[100001];
priority_queue<long long,vector<long long>,greater<long long> > Q;
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	int i;
	for(i=1;i<=n;i++)
	{
		scanf("%d",&a[i].x);
		a[i].p=i-1;
	}
	for(i=1;i<=n;i++)
		fa[i]=i;
	for(i=1;i<=m;i++)
	{
		int s,t;
		scanf("%d%d",&s,&t);
		int fx=find(s),fy=find(t);
		if(fx!=fy)
			fa[fx]=fy;
	}
	sort(a+1,a+1+n);
	long long ans=0;
	int sum=0;
	for(i=1;i<=n;i++)
	{
		int fx=find(a[i].p);
		if(!v[fx])
		{
			sum++;
			v[fx]=true;
			ans+=a[i].x;
		}
		else
			Q.push(a[i].x);
	}
	int xx=0;
	while(!Q.empty()&&xx<sum-2)
	{
		xx++;
		ans+=Q.top();
		Q.pop(); 
	}
	if(sum==1)
		printf("0\n");
	else if(xx<sum-2)
		printf("Impossible\n");
	else
		printf("%lld\n",ans);
	return 0;
}