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 <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <iomanip>
#define MOD 1000000007
typedef long long ll;
using namespace std;

vector<int> g[100010];
int col[100010];
ll a[100010];
int mi,mit;

void dfs(int now,int c){
  col[now]=c;
  if(mi>a[now]){
    mi=a[now];
    mit=now;
  }
  for(int i=0;i<g[now].size();i++){
    int u=g[now][i];
    if(col[u]!=0) continue;
    dfs(u,c);
  }
  return;
}

int main(){
  int n,m;
  cin>>n>>m;
  for(int i=0;i<n;i++){
    cin>>a[i];
  }
  for(int i=0;i<m;i++){
    int x,y;
    cin>>x>>y;
    g[x].push_back(y);
    g[y].push_back(x);
  }

  if(n<2*(n-m-1)){
    cout<<"Impossible"<<endl;
    return 0;
  }else if(m==n-1){
    cout<<0<<endl;
    return 0;
  }

  int cnt=1;
  ll ans=0;
  for(int i=0;i<n;i++){
    if(col[i]==0){
      mi=1e9+10;
      mit=0;
      dfs(i,cnt);
      cnt++;
      ans=ans+mi;
      col[mit]=-1;
    }
  }

  vector<ll> v;
  for(int i=0;i<n;i++){
    if(col[i]!=-1) v.push_back(a[i]);
  }
  sort(v.begin(),v.end());
  for(int i=0;i<v.size();i++){
    if(cnt-1>=2*(n-m-1)) break;
    ans=ans+v[i];
    cnt++;
  }

  cout<<ans<<endl;

  return 0;
}
