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<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cassert>

typedef long long ll;
typedef unsigned long long ull;

using namespace std;

const int MAXN=1E5+10;

int n,m,a[MAXN];
vector<int> g[MAXN],f[MAXN],glo;
bool vis[MAXN];
ll Ans;

void dfs(int x,int fa,vector<int> &ve)
{
	ve.push_back(a[x]);
	vis[x]=1;
	for(int v:g[x])
		if(v!=fa)
			dfs(v,x,ve);
}

int main()
{
	scanf("%d%d",&n,&m);
	if(m==n-1) return puts("0"),0;
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=1,u,v;i<=m;i++)
		scanf("%d%d",&u,&v),++u,++v,g[u].push_back(v),g[v].push_back(u);
	int cnt=0;
	for(int i=1;i<=n;i++)
		if(!vis[i])
		{
			++cnt;
			dfs(i,0,f[i]);
			sort(f[i].begin(),f[i].end());
			Ans+=f[i][0];
			for(int j=1;j<f[i].size();j++)
				glo.push_back(f[i][j]);
		}
	if(((cnt-1)<<1)>n) return puts("Impossible"),0;
	sort(glo.begin(),glo.end());
	for(int i=0;i<cnt-2;i++)
		Ans+=glo[i];
	printf("%lld\n",Ans);
}