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;

int N, M;
#define MAX_N 100000
long long A[MAX_N];
int R[MAX_N], P[MAX_N];
map<int, vector<long long>> S;
vector<long long> X;
long long Z;

int find(int v) {
  if(P[v] == v) return v;
  return P[v] = find(P[v]);
}
void unite(int v, int w) {
  v = find(v); w = find(w);
  if(v == w) return;
  if(R[v] > R[w]) swap(v, w);
  P[v] = w; if(R[v] == R[w]) ++R[w];
}
int main() {
  cin >> N >> M;
  for(int v = 0; v < N; ++v) {
    cin >> A[v]; P[v] = v;
  }
  for(int i = 0; i < M; ++i) {
    int x, y; cin >> x >> y; unite(x, y);
  }
  for(int v = 0; v < N; ++v) S[find(v)].push_back(A[v]);
  int K = S.size();
  if(K == 1) { cout << 0 << endl; return 0; }
  for(auto kv : S) {
    vector<long long>& a = kv.second;
    sort(a.begin(), a.end());
    Z += a[0];
    for(int i = 1; i < (int)a.size(); ++i) X.push_back(a[i]);
  }
  if((int)X.size() < K - 2) { cout << "Impossible" << endl; return 0; }
  sort(X.begin(), X.end());
  for(int i = 0; i < K - 2; ++i) Z += X[i];
  cout << Z << endl;
  return 0;
}
