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;

int memo[100005];
set<int> edge[100005];

void dfs(int now)
{
	for (auto v:edge[now])
	{
		if (memo[v]==0)
		{
			memo[v]=memo[now];
			dfs(v);
		}
	}
}

int main(int argc, char const *argv[])
{
	int n,m;
	cin>>n>>m;
	int a[n];
	for (int i = 0; i < n; ++i)
	{
		cin>>a[i];
	}
	int x,y;
	for (int i = 0; i < m; ++i)
	{
		cin>>x>>y;
		edge[x].insert(y);
		edge[y].insert(x);
	}
	if (m+n/2<n-1)
	{
		cout<<"Impossible\n";
		return 0;
	}
	int cnt=0;
	for (int i = 0; i < n; ++i)
	{
		if (memo[i]==0)
		{
			++cnt;
			memo[i]=cnt;
			dfs(i);
		}
	}
	if (cnt==1)
	{
		cout<<"0\n";
		return 0;
	}
	long long mi[cnt+5];
	for (int i = 0; i < cnt+5; ++i)
	{
		mi[i]=(1LL<<50);
	}
	multiset<long long> r;
	for (int i = 0; i < n; ++i)
	{
		if (mi[memo[i]]==(1LL<<50))
		{
			mi[memo[i]]=a[i];
		}
		else if (mi[memo[i]]>a[i])
		{
			r.insert(mi[memo[i]]);
			mi[memo[i]]=a[i];
		}
		else
		{
			r.insert(a[i]);
		}
	}
	long long ans=0LL;
	for (int i = 1; i <= cnt; ++i)
	{
		ans+=mi[i];
	}
	cnt-=2;
	for (auto v:r)
	{
		if (cnt<=0)
		{
			break;
		}
		ans+=v;
		--cnt;
	}
	cout<<ans<<endl;
	return 0;
}
