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
// I SELL YOU...! 
#include<iostream>
#include<vector>
#include<algorithm>
#include<functional>
#include<queue>
#define MAX_N 100010
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
vector<ll> G[MAX_N];
ll a[MAX_N];
bool used[MAX_N]={};
void dfs(int v,vector<ll>& vals){
  if(used[v]) return;
  used[v] = true;
  vals.push_back(a[v]);
  for(auto i:G[v]){
    dfs(i,vals);
  }
}
signed main(){
  ll n,m,ans=0,need;
  cin >> n >> m;
  ll x[m],y[m];
  priority_queue<ll,vector<ll>,greater<ll>> que;
  for(int i=0;i<n;i++){
    cin >> a[i];
  }
  for(int i=0;i<m;i++){
    cin >> x[i] >> y[i];
    G[x[i]].push_back(y[i]);
    G[y[i]].push_back(x[i]);
  }
  if(m==n-1){
    cout <<"0\n";
    return 0;
  }
  need = 2*(n-m-1);
  for(int i=0;i<n;i++){
    vector<ll> vt;
    if(!used[i]){
      dfs(i,vt);
      sort(vt.begin(),vt.end());
      ans += vt[0];
      need--;
      for(int j=1;j<vt.size();j++){
        que.push(vt[j]);
      }
    }
  }
  if(need<0){
    cout <<"Impossible\n";
    return 0;
  }
  for(int i=0;i<need;i++){
    if(que.empty()){
      cout <<"Impossible\n";
      return 0;
    }
    ans += que.top();
    que.pop();
  }
  cout << ans << endl;
}
