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 <cstdio>
#include <algorithm>
#include <numeric>
#include <functional>
#include <vector>
using namespace std;

const int kMaxN = 1e5 + 5;

typedef long long ll;

int n, m;
ll a[kMaxN];
vector<int> G[kMaxN], S;
bool vis[kMaxN];

int DFS(int x, vector<int> &vec) {
  vec.push_back(a[x]);
  vis[x] = true;
  for (int nxt : G[x])
    if (!vis[nxt])
      DFS(nxt, vec);
}

int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i)
    scanf("%d", a + i);
  for (int i = 0, x, y; i < m; ++i) {
    scanf("%d%d", &x, &y);
    G[x].push_back(y);
    G[y].push_back(x);
  }
  if (m == n - 1)
    return puts("0"), 0;
  int need = (n - m - 1) << 1;
  if (n < need) {
    return puts("Impossible"), 0;
  }
  ll ans = 0;
  for (int i = 0; i < n; ++i)
    if (!vis[i]) {
      --need;
      vector<int> v;
      DFS(i, v);
      sort(v.begin(), v.end());
      ans += v[0];
      for (int i = 1; i < v.size(); ++i)
        S.push_back(v[i]);
    }
  sort(S.begin(), S.end());
  for (int i = 0; i < need; ++i)
    ans += S[i];
  printf("%lld\n", ans);
  return 0;
}