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;

typedef long long LL;

#define pb push_back

const int MAXN=100005;

int a[MAXN],cnt,b[MAXN],blo,c[MAXN];
bool vis[MAXN];
vector <int> G[MAXN];
queue <int> q;
LL ans;

void bfs(int u)
{
	blo++;
	int mn=a[u];
	vis[u]=true;
	q.push(u);
	int p=0;
	while (!q.empty())
	{
		int u=q.front();q.pop();
		c[p++]=a[u];
		for (int v:G[u])
			if (!vis[v])
			{
				vis[v]=true;
				mn=min(mn,a[v]);
				q.push(v);
			}
	}
	ans+=mn;
	for (int i=0;i<p;i++)
		if (c[i]!=mn)
			b[cnt++]=c[i];
		else
			mn=-1;
}

int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	for (int i=0;i<n;i++)
		scanf("%d",&a[i]);
	for (int i=0;i<m;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		G[u].pb(v);
		G[v].pb(u);
	}
	for (int i=0;i<n;i++)
		if (!vis[i])
			bfs(i);
	if (blo==1)
		printf("0");
	else if (n<2*(blo-1))
		printf("Impossible");
	else
	{
		nth_element(b,b+blo-2,b+cnt);
		for (int i=0;i<blo-2;i++)
			ans+=b[i];
		printf("%lld",ans);
	}
	return 0;
}