Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

7 5
1 2 3 4 5 6 7
3 0
4 0
1 2
1 3
5 6

Sample Output 1

7

If we connect vertices 0 and 5, the graph becomes connected, for the cost of 1 + 6 = 7 dollars.

Sample Input 2

5 0
3 1 4 1 5

Sample Output 2

Impossible

We can't make the graph connected.

Sample Input 3

1 0
5

Sample Output 3

0

The graph is already connected, so we do not need to add any edges.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define int long long
#define rep(i,l,r) for(int i=(int)(l);i<(int)(r);i++)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)x.size())
template<class T>bool chmax(T &a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T>bool chmin(T &a,T b){if(a>b){a=b;return 1;}return 0;}

/*
*/

using vi = vector<int>;
using vvi = vector<vi>;
using P = pair<int,int>;

class UnionFind
{
public:
  int n;
  vector<int> par, rnk;

  UnionFind(int n_)
  {
    n = n_;
    par.resize(n);
    rnk.resize(n);
    for(int i = 0; i < n; i++){
      par[i] = i;
      rnk[i] = 0;
    }
  }

  int find(int x)
  {
    if(par[x] == x) return x;
    return par[x] = find(par[x]);
  }

  void unite(int x, int y)
  {
    x = find(x);
    y = find(y);
    if(x == y) return;

    if(rnk[x] < rnk[y]){
      par[x] = y;
    }else{
      par[y] = x; 
      if(rnk[x] == rnk[y]) rnk[x]++;
    }
  }

  bool same(int x, int y)
  {
    return find(x) == find(y);
  }
};

signed main() {
  int n, m;
  cin >> n >> m;

  vi a(n);
  rep(i, 0, n) cin >> a[i];

  UnionFind uf(n);
  rep(i, 0, m){
    int x, y;
    cin >> x >> y;
    uf.unite(x, y);
  }

  set<int> st;
  rep(i, 0, n) st.insert(uf.find(i));

  int siz = sz(st);
  if(siz == 1){
    cout << 0 << endl;
  }else{
    int ans = 0;
    priority_queue<int, vi, greater<int>> que[n];
    rep(i, 0, n) que[uf.find(i)].push(a[i]);
    rep(i, 0, n){
      if(sz(que[i]) == 0) continue;
      ans += que[i].top();
      que[i].pop();
    }

    priority_queue<int, vi, greater<int>> que2;
    rep(i, 0, n){
      while(!que[i].empty()){
        que2.push(que[i].top());
        que[i].pop();
      }
    }

    if(sz(que2) < siz-2){
      cout << "Impossible" << endl;
    }else{
      rep(i, 0, siz-2){
        ans += que2.top();
        que2.pop();
      }
      
      cout << ans << endl;
    }
  }

  return 0;
}