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 <cmath>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <vector>
using namespace std;
using ll = int64_t;
#define rep(i, j, n) for (int i = j; i < (int)n; ++i)
#define rrep(i, j, n) for (int i = (int)n - 1; j <= i; --i)

int main() {
  int n, m;
  cin >> n >> m;
  vector<ll> a(n);
  for (ll& x : a) cin >> x;
  vector<vector<int>> g(n);
  rep(i, 0, m) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }

  // 木の数がkだったら、辺はk-1個必要
  // 頂点2つで辺を1本生やせるので、 (k - 1) * 2 > nだったら不可能
  vector<priority_queue<ll, vector<ll>, greater<ll>>> vert(n);
  vector<bool> used(n);
  auto dfs = [&](auto self, int v, int cnt) -> void {
    vert[cnt].emplace(a[v]);
    used[v] = true;
    for (int nv : g[v])
      if (!used[nv]) self(self, nv, cnt);
  };
  int cnt = 0;
  rep(i, 0, n) {
    if (!used[i]) {
      dfs(dfs, i, cnt);
      ++cnt;
    }
  }
  if ((cnt - 1) * 2 > n) {
    cout << "Impossible" << endl;
  } else {
    // (k - 1) * 2個、各成分から
    ll ans = 0;
    int need = (cnt - 1) * 2;
    while (need > 0) {
      vector<ll> tmp;
      rep(i, 0, cnt) {
        if (vert[i].size()) {
          ll v = vert[i].top();
          tmp.emplace_back(v);
          vert[i].pop();
        }
      }
      if (need >= (int)tmp.size()) {
        rep(i, 0, tmp.size()) ans += tmp[i];
        need -= tmp.size();
      } else {
        sort(tmp.begin(), tmp.end());
        rep(i, 0, need) ans += tmp[i];
        need = 0;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}