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<queue>
using namespace std;
int anc[101010];
int a[101010];
void find(int x)
{
	if(anc[anc[x]]!=anc[x])
	{
		find(anc[x]);
		anc[x]=anc[anc[x]];
	}
}
int cnt;
void link(int u,int v)
{
	find(u);
	find(v);
	if(anc[u]!=anc[v])
	{
		cnt--;
		anc[anc[v]]=anc[u];
	}
}
struct Node{
	Node(int _anc,int _val)
	{
		anc=_anc;
		val=_val;
	}
	int anc,val;
};
bool vis[101010];
bool operator < (const Node &A,const Node &B)
{
	return A.val>B.val;
}
priority_queue<Node> Q;
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	cnt=n;
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=1;i<=n;i++)
		anc[i]=i;
	for(int i=1;i<=m;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		u++;
		v++;
		link(u,v);
	}
	for(int i=1;i<=n;i++)
	{
		find(i);
		Q.push(Node(anc[i],a[i]));
	}
	long long ans=0;
	for(int i=2*cnt-2;i>=1;i--)
	{
		if(Q.empty())
		{
			printf("Impossible\n");
			return 0;
		}
		while(vis[Q.top().anc]!=0&&i<=cnt)
		{
			Q.pop();
		}
		if(vis[Q.top().anc]==0)
		{
			cnt--;
			vis[Q.top().anc]=1;
		}
		ans+=Q.top().val;
		Q.pop();
	}
	printf("%lld\n",ans);
	return 0;
}