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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef pair<int, int> Pii;
typedef pair<int, ll> Pil;
typedef pair<ll, ll> Pll;
typedef pair<ll, int> Pli;

#define fi first
#define se second

const ll MOD = 1e9 + 7;
const ll MOD2 = 998244353;
const ll MOD3 = 1812447359;
const ll INF = 1ll << 62;
const double PI = 2 * asin(1);

void yes() {printf("yes\n");}
void no() {printf("no\n");}
void Yes() {printf("Yes\n");}
void No() {printf("No\n");}
void YES() {printf("YES\n");}
void NO() {printf("NO\n");}

int N, M, G[int(1e5+5)];
ll A[int(1e5+5)];
vector <ll> now[int(1e5+5)];

void Unite(){
  int X, Y; cin >> X >> Y;
  int GX = G[X], GY = G[Y];
  while (GX != G[GX]) GX = G[GX];
  while (GY != G[GY]) GY = G[GY];

  G[max(GX, GY)] = min(GX, GY);
}

int main(){
  cin >> N >> M;
  for (int i = 0; i < N; i++) {
    cin >> A[i]; G[i] = i;
  }
  for (int i = 0; i < M; i++) Unite();
  for (int i = 0; i < N; i++){
    int Gi = G[i];
    while (Gi != G[Gi]) Gi = G[Gi];
    G[i] = Gi;
    now[G[i]].push_back(A[i]);
  }

  priority_queue <ll, vector<ll>, greater<ll> > que;
  for (int i = 0; i < (int)now[0].size(); i++) que.push(now[0][i]);
  
  ll ans = 0;
  for (int i = 1; i < N; i++){
    if (now[i].size() == 0) continue;

    if (que.size() == 0){
      cout << "Impossible" << endl;
      return 0;
    }
    ans += que.top(); que.pop();
    sort(now[i].begin(), now[i].end());
    ans += now[i][0];

    for (int j = 1; j < now[i].size(); j++){
      que.push(now[i][j]);
    }
  }
  cout << ans << endl;

  return 0;
}