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<cstring>
#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;
const int maxm=100007;
int n,m;
int tot;//连通块个数 
int w[maxm],f[maxm];
int cnt[maxm];
vector<int> t,q[maxm];
ll ans=0;
int find(int x)
{
 if(x!=f[x])
 f[x]=find(f[x]);
 return f[x];	
}
int main()
{
 scanf("%d%d",&n,&m);
 for(int i=1;i<=n;i++)
 scanf("%d",&w[i]),f[i]=i;
 for(int i=1;i<=m;i++)
 {
  int x,y;
  scanf("%d%d",&x,&y);
  x++;//编号初始为0，++变1 
  y++;
  int fx=find(x);
  int fy=find(y);
  if(fx!=fy) f[fx]=fy;
 }
 for(int i=1;i<=n;i++)
 {
   int fx=find(i);
   if(!cnt[fx]) cnt[fx]=++tot;
   q[cnt[fx]].push_back(w[i]);
 }
 if(tot==1)
 {
  printf("0\n");
  return 0; 
 } 
 for(int i=1;i<=tot;i++)
 {
  sort(q[i].begin(),q[i].end());
  ans+=q[i][0];
  for(int j=1;j<q[i].size();j++)
  {
  	t.push_back(q[i][j]);
  }
 }
 if(t.size()<tot-2)
 {
  printf("Impossible\n");
  return 0;	
 }
 sort(t.begin(),t.end());
 for(int i=0;i<tot-2;i++)
 ans+=t[i];
 printf("%lld\n",ans);
 return 0;	
}
