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>
#define maxl 100010
using namespace std;

int n,m;
pair<int,int>p[maxl];
int a[maxl],f[maxl];
bool vis[maxl];
int cnt,res;
long long ans;
inline int find(int x)
{
	if(f[x]!=x)
		f[x]=find(f[x]);
	return f[x];
}

inline void prework()
{
	for(int i=1;i<=n;i++)
	{
		f[i]=i;
	}
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	int u,v,x,y;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d",&u,&v);
		u++;v++;
		x=find(u);y=find(v);
		if(x!=y)
			f[y]=x;
	}
	for(int i=1;i<=n;i++)
		x=find(i);
	for(int i=1;i<=n;i++)
		p[i]=make_pair(a[i],f[i]);
	sort(p+1,p+1+n);
}

inline void mainwork()
{
	cnt=n-m,res=0;
	ans=0;
	for(int i=1;i<=n;++i)
		if(!vis[p[i].second])
		{
			--cnt;
			ans+=p[i].first;
			++res;
			vis[p[i].second]=1;
		}
		else
		{
			if(res+cnt<(n-m-1)*2)
				ans+=p[i].first,++res;
		}
}

inline void print()
{
	if((n-m-1)==0)printf("0\n");
	else if(cnt==0&&res==(n-1-m)*2)
		printf("%lld\n",ans);
	else
		puts("Impossible");
}

int main()
{
	while(~scanf("%d%d",&n,&m))
	{
		prework();
		mainwork();
		print();
	}
	return 0;
}
 