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;

const int MAX_N = 100000;
vector<int> graph[MAX_N];

int costnod[MAX_N];
bool viz[MAX_N];

int dfs(int nod) {
  int minnod = nod;
  viz[nod] = true;
  for(auto it:graph[nod])
    if(!viz[it]) {
      int r = dfs(it);
      if(costnod[r] < costnod[minnod])
        minnod = r;
    }
 return minnod;
}

int main() {
  int n, m, a, b, comp, x;
  long long rez = 0LL;
  scanf("%d%d", &n, &m);
  for(int i = 0; i < n; ++i)
    scanf("%d", &costnod[i]);
  for(int i = 0; i < m; ++i) {
    scanf("%d%d", &a, &b);
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  
  comp = 0;
  for(int i = 0; i < n; ++i)
    if(!viz[i]) {
      comp++;
      x = dfs(i);
      rez += costnod[x];
      costnod[x] = 1000000001;
    }
  if(comp > 1) {
    if(n < 2 * (comp - 1))
      printf("Impossible");
    else {
      std::sort(costnod, costnod + n);
      for(int i = 0; i < comp - 2; ++i)
        rez += costnod[i];
      printf("%lld", rez);
    }
  } else if(comp == 1)
    printf("0");
  return 0;
}
