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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

from heapq import heapify, heappop, heappush

class UnionFind(object):
    def __init__(self, n=1):
        self.par = [i for i in range(n)]
        self.rank = [0 for _ in range(n)]
        self.size = [1 for _ in range(n)]
    def find(self, x):
        """
        x が属するグループを探索
        """
        if self.par[x] == x:
            return x
        else:
            self.par[x] = self.find(self.par[x])
            return self.par[x]
    def union(self, x, y):
        """
        x と y のグループを結合
        """
        x = self.find(x)
        y = self.find(y)
        if x != y:
            if self.rank[x] < self.rank[y]:
                x, y = y, x
            if self.rank[x] == self.rank[y]:
                self.rank[x] += 1
            self.par[y] = x
            self.size[x] += self.size[y]
    def is_same(self, x, y):
        """
        x と y が同じグループか否か
        """
        return self.find(x) == self.find(y)
    def get_size(self, x):
        """
        x が属するグループの要素数
        """
        x = self.find(x)
        return self.size[x]

N,M = map(int,input().split())
A = list(map(int,input().split()))
Cost = []
for i in range(N):
  heappush(Cost,[A[i],i]) #コスト、IDという順で保存。
uf = UnionFind(N)
for i in range(M):
  a,b = map(int,input().split())
  #a-=1;b-=1 #既に0index
  uf.union(a,b)
ans = 0
f_cost,first = heappop(Cost)
Nakama = []
heappush(Nakama,[f_cost,first])
while Cost and Nakama and M < N-1: #辺の数がN-1よりも小さいなら
  n_cost,nxt = heappop(Cost)
  if uf.is_same(first, nxt): #同じグループなら仲間に追加
    heappush(Nakama,[n_cost,nxt])
  else: #違うグループならこいつと仲間の最小を出して足す。
    nakama_cost,nakama = heappop(Nakama)
    print(n_cost,nxt,nakama_cost,nakama)
    ans += n_cost + nakama_cost #コストの足し算
    uf.union(nxt,nakama) #nxtを仲間に入れる。
    M += 1 #辺の数を足す。
if M < N-1:
  print("Impossible")
else:
  print(ans)