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 Python (3.4.3) (judge verdict: Wrong Answer)

class UnionFind():
   # 作りたい要素数nで初期化
   # 使用するインスタンス変数の初期化
   def __init__(self, n):
       self.n = n
       # root[x]<0ならそのノードが根かつその値が木の要素数
       # rootノードでその木の要素数を記録する
       self.root = [-1]*(n+1)
       # 木をくっつける時にアンバランスにならないように調整する
       self.rnk = [0]*(n+1)
   # ノードxのrootノードを見つける
   def Find_Root(self, x):
       if(self.root[x] < 0):
           return x
       else:
           # ここで代入しておくことで、後の繰り返しを避ける
           self.root[x] = self.Find_Root(self.root[x])
           return self.root[x]
   # 木の併合、入力は併合したい各ノード
   def Unite(self, x, y):
       # 入力ノードのrootノードを見つける
       x = self.Find_Root(x)
       y = self.Find_Root(y)
       # すでに同じ木に属していた場合
       if(x == y):
           return
       # 違う木に属していた場合rnkを見てくっつける方を決める
       elif(self.rnk[x] > self.rnk[y]):
           self.root[x] += self.root[y]
           self.root[y] = x
       else:
           self.root[y] += self.root[x]
           self.root[x] = y
           # rnkが同じ（深さに差がない場合）は1増やす
           if(self.rnk[x] == self.rnk[y]):
               self.rnk[y] += 1
   # xとyが同じグループに属するか判断
   def isSameGroup(self, x, y):
       return self.Find_Root(x) == self.Find_Root(y)
   # ノードxが属する木のサイズを返す
   def Count(self, x):
       return -self.root[self.Find_Root(x)]
n,m=map(int,input().split())
a=list(map(int,input().split()))
h=UnionFind(n)
for i in range(m):
    x,y=map(int,input().split())
    h.Unite(x,y)
dic={}
k=0
for i in range(n):
    x=h.Find_Root(i)
    if x in dic:
        dic[x].add(i)
    else:
        dic[x]={i}
        k+=1
b=[sorted([a[j] for j in dic[i]]) for i in dic]
ans=0
c=[]
cnt=0
if k==1:
    print(0)
    exit()
for i in range(k):
    ans+=b[i][0]
    if len(b[i])>1:
        for j in b[i][1:]:
            c.append(j)
    else:
        cnt+=1
if k==2:
    print(ans)
    exit()
if cnt>2:
    print("Impossible")
    exit()
else:
    c.sort()
    ans+=sum(c[:k-2])
print(ans)