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;


#define MAXN 100001
typedef long long ll;

int n, m, a[MAXN];
vector<int> g[MAXN];
int bio[MAXN];
set<pair<int, int>> S;

pair<int, int> dfs(int u) {
  bio[u] = 1;
  pair<int, int> ret = {a[u], u};
  for (auto &v : g[u]) {
    if (bio[v]) continue;
    ret = min(ret, dfs(v));
  }
  return ret;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  for (int u = 0; u < n; ++u) {
    cin >> a[u];
    S.insert({a[u], u});
  }
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  int N = 0;
  ll ans = 0;
  for (int u = 0; u < n; ++u) {
    if (bio[u]) continue;
    auto cmn = dfs(u);
    S.erase(cmn);
    ans += (ll)cmn.first;
    N++;
  }
  if (N==1) {
    cout << 0 << endl;
    return 0;
  }
  if (2*N-2>n) {
    cout << "Impossible" << endl;
    return 0;
  }
  auto it = S.begin();
  for (int i = 0; i < N-2; ++i, ++it) {
    ans += (ll)it->first;
  }
  cout << ans << endl;
  return 0;
}
