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 <algorithm>
#include <cassert>
#include <iostream>
#include <vector>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
vector<int> parent;
vector<int> nodes;

int getParent(int x) {
  int y = x;
  while (y != parent[y]) {
    y = parent[y];
  }
  parent[x] = y;
  return y;
}

bool unite(int x, int y) {
  x = getParent(x);
  y = getParent(y);
  if (x == y) return false;
  if (nodes[x] < nodes[y]) swap(x, y);
  parent[y] = x;
  nodes[x] += nodes[y];
  return true;
}

int main(void) {
  ios::sync_with_stdio(false);
  int N, M;
  cin >> N >> M;
  vector<pair<int64_t, int>> A(N);
  rep(i, N) {
    int64_t a;
    cin >> a;
    A[i] = {a, i};
  }

  parent.resize(N);
  nodes.resize(N);
  rep(i, N) {
    parent[i] = i;
    nodes[i] = 1;
  }
  int pnum = N;
  rep(i, M) {
    int x, y;
    cin >> x >> y;
    if (unite(x, y)) pnum--;
  }
  int need = 2 * (pnum - 1);
  if (N < need) {
    cout << "Impossible" << endl;
    return 0;
  }

  sort(A.begin(), A.end());
  vector<bool> usedP(N, false);
  vector<bool> usedA(N, false);
  int64_t answer = 0;
  int count = 0;
  rep(i, N) {
    int x = getParent(i);
    if (usedP[x]) continue;
    usedP[x] = true;
    usedA[i] = true;
    answer += A[i].first;
    count++;
  }
  if (count == 1) answer = 0;
  if (N < (count - 1) * 2) {
    cout << answer << endl;
    return 0;
  }
  count = (count - 1) * 2 - count;
  rep(i, N) {
    if (usedA[i]) continue;
    if (count == 0) break;
    answer += A[i].first;
    count--;
  }
  cout << answer << endl;
  return 0;
}