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
#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> g, mn;
vector<int> a, vis;

void dfs(int x, int id) {
  vis[x] = 1;
  mn[id].push_back(a[x]);
  for (auto v : g[x]) {
    if (vis[v]) continue;
    dfs(v, id);
  }
}

int main() {

  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);

  int n, m;
  cin >> n >> m;

  a.resize(n);
  g.resize(n);
  mn.resize(n - m);
  vis.resize(n, 0);

  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }

  if (n < 2 * (n - m - 1)) {
    cout << "Impossible\n";
    return 0;
  }
  if (m == n-1) {
    cout << 0 << '\n';
    return 0;
  }

  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;

    g[u].push_back(v);
    g[v].push_back(u);
  }

  long long ans = 0;
  for (int i = 0, cur = 0; i < n; ++i) {
    if (!vis[i]) {
      dfs(i, cur++);
    }
  }

  vector<int> rem; 
  for (int i = 0; i < (int)mn.size(); ++i) {
    sort(mn[i].begin(), mn[i].end());
    for (int j = 1; j < (int)mn[i].size(); ++j) {
      rem.push_back(mn[i][j]);
    }
    ans += mn[i][0];
  }

  sort(rem.begin(), rem.end());
  for (int i = 0; i < n - m - 2; ++i) {
    ans += rem[i];
  }

  cout << ans << '\n';
}
