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;

int main()
{
  int n,m;
  scanf("%d%d",&n,&m);
  if(m==n-1){
    printf("0\n");
    return 0;
  }
  static vector<ll> a(n);
  for(int i=0;i<n;i++){
    scanf("%lld",&(a[i]));
  }
  vector<vector<int> > e(n);
  for(int t=0;t<m;t++){
    int x,y;
    scanf("%d%d",&x,&y);
    e[x].push_back(y);
    e[y].push_back(x);
  }
  vector<int> vis(n,0);
  vector<vector<ll> > A;
  for(int i=0;i<n;i++){
    if(vis[i])continue;
    vis[i]=1;
    vector<ll> c(1,a[i]);
    queue<int> Q;
    Q.push(i);
    while(!Q.empty()){
      int j=Q.front();Q.pop();
      for(int x=0;x<e[j].size();x++){
	int k=e[j][x];
	if(!vis[k]){
	  vis[k]=1;
	  Q.push(k);
	  c.push_back(a[k]);
	}
      }
    }
    sort(c.begin(),c.end());
    A.push_back(c);
  }
  int K=A.size();
  long long ans=0;
  vector<int> b;
  for(int t=0;t<K;t++){
    ans+=A[t][0];
    for(int x=1;x<A[t].size();x++){
      b.push_back(A[t][x]);
    }
  }
  if(b.size()<K-2){
    puts("Impossible");
  }
  else{
    sort(b.begin(),b.end());
    for(int t=0;t<K-2;t++){
      ans+=b[t];
    }
    printf("%lld\n",ans);
  }
  return 0;
}
