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;

typedef long long int LLI;
typedef priority_queue<LLI, vector<LLI>, greater<LLI>> Q;

int N;
int M;
LLI ans;
LLI as[114514];
bool used[114514];
Q que;
Q qs[114514];
vector<int> es[114514];

void dfs(int v, Q &q) {
  used[v] = true;
  q.push(as[v]);
  for (int u : es[v]) {
    if (used[u]) continue;
    dfs(u, q);
  }
}

int main() {
  scanf("%d%d", &N, &M);
  for (int i=0; i<N; i++) scanf("%lld", &as[i]);
  for (int i=0; i<M; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    es[a].emplace_back(b);
    es[b].emplace_back(a);
  }

  int num = 0;
  for (int i=0; i<N; i++) {
    if (!used[i]) {
      dfs(i, qs[i]);
      ans += qs[i].top(); qs[i].pop();
      if (qs[i].size() > que.size()) swap(qs[i], que);
      while (!qs[i].empty()) {
        LLI v = qs[i].top(); qs[i].pop();
        que.push(v);
      }
      num++;
    }
  }

  if (que.size()+2 < num) {
    puts("Impossible");
    return 0;
  }

  for (int i=0; i<num-2; i++) {
    LLI v = que.top(); que.pop();
    ans += v;
  }

  if (num == 1) puts("0");
  else printf("%lld\n", ans);  
}
