Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

7 5
1 2 3 4 5 6 7
3 0
4 0
1 2
1 3
5 6

Sample Output 1

7

If we connect vertices 0 and 5, the graph becomes connected, for the cost of 1 + 6 = 7 dollars.

Sample Input 2

5 0
3 1 4 1 5

Sample Output 2

Impossible

We can't make the graph connected.

Sample Input 3

1 0
5

Sample Output 3

0

The graph is already connected, so we do not need to add any edges.

Solution in Ruby (2.3.3) (judge verdict: Accepted)

class UnionFind
  def initialize(n)
    @size = Array.new(n + 1, 1)
    @rank = Array.new(n + 1, 0)
    @parent = []

    (0..n).each do |i|
      @parent[i] = i
    end
  end

  def find(x)
    if @parent[x] == x
      x
    else
      @parent[x] = find(@parent[x])
    end
  end

  def unite(x, y)
    x = find(x)
    y = find(y)
    return if x == y

    if @rank[x] < @rank[y]
      @parent[x] = y
      @size[y] += @size[x]
    else
      @parent[y] = x
      @size[x] += @size[y]

      @rank[x] += 1 if @rank[x] == @rank[y]
    end
  end

  def same?(x, y)
    find(x) == find(y)
  end

  def size(x)
    @size[find(x)]
  end
end

class PriorityQueue
  attr_reader :size

  def initialize(order: :asc)
    @size = 0
    @nodes = []
  end

  def empty?
    @size.zero?
  end

  def push(node)
    index = size
    @size += 1

    while index > 0
      parent = (index - 1) / 2

      break if (@nodes[parent] <=> node) <= 0

      @nodes[index] = @nodes[parent]
      index = parent
    end

    @nodes[index] = node
  end

  alias_method :<<, :push

  def pop
    node = @nodes.first
    root = @nodes[size - 1]
    @size -= 1
    index = 0

    while index * 2 + 1 < @size
      left = 2 * index + 1
      right = left + 1

      if right < size && (@nodes[right] <=> @nodes[left]) <= 0
        left = right
      end

      break if (@nodes[left] <=> root) > 0

      @nodes[index] = @nodes[left]
      index = left
    end

    @nodes[index] = root
    node
  end

  def top
    @nodes.first
  end
end

Node = Struct.new(:v, :cost) do
  def <=>(node)
    cost <=> node.cost
  end
end

N, M = gets.split.map(&:to_i)
A = gets.split.map(&:to_i)

uf = UnionFind.new(N + 1)

M.times do
  x, y = gets.split.map(&:to_i)

  uf.unite(x, y)
end

trees = Hash.new { |h, k| h[k] = [] }

N.times do |v|
  trees[uf.find(v)] << v
end

trees = trees.sort_by { |_, list| -list.size }

base = trees.shift
ans = 0

pque = PriorityQueue.new

base.last.each do |v|
  pque.push(Node.new(v, A[v]))
end

while !trees.empty? && !pque.empty?
  tree = trees.shift
  stock = PriorityQueue.new

  tree.last.each do |v|
    stock.push(Node.new(v, A[v]))
  end

  v = pque.pop
  u = stock.pop

  ans += v.cost + u.cost

  until stock.empty?
    pque.push(stock.pop)
  end
end

if trees.empty?
  puts ans
else
  puts 'Impossible'
end