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<vector>
#include<queue>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define int long long
const int N=100000+10;

int _r;char _c;

int read()
{
	while(!isdigit(_c=getchar()));
	_r=_c-48;
	while(isdigit(_c=getchar()))
		_r=_r*10-48+_c;
	return _r;
}

int n,m,a[N],fa[N],cnt,ans;
vector<int> v[N];
priority_queue<int,vector<int>,greater<int> > Q;

int find(int x){return (fa[x]==x)?x:(fa[x]=find(fa[x]));}

void hb(int x,int y)
{
	x=find(x);y=find(y);
	if(x!=y)
		fa[x]=y;
}

signed main()
{
	n=read();m=read();
	int i;
	for(i=1;i<=n;++i)
		a[fa[i]=i]=read();
	while(m--)
		hb(read()+1,read()+1);
	for(i=1;i<=n;++i)
		v[find(i)].push_back(a[i]);
	vector<int>::iterator it;
	for(i=1;i<=n;++i)if(fa[i]==i)
	{
		++cnt;
		sort(v[i].begin(),v[i].end());
		ans+=*v[i].begin();
		for(it=v[i].begin()+1;it!=v[i].end();++it)
			Q.push(*it);
	}
	if(cnt==1)
		printf("0");
	else if(cnt-2>Q.size())
		printf("Impossible");
	else
	{
		while(cnt-->2)
		{
			ans+=Q.top();
			Q.pop();
		}
		printf("%lld",ans);
	}        	
	return 0;
}
