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
n,m=map(int,input().split())

a=list(map(int,input().split()))

union=[-1]*n

tree=[[] for _ in range(n)]
for _ in range(m):
    x,y=map(int,input().split())
    tree[x].append(y)
    tree[y].append(x)

cnt=-1
for i in range(n):
    if union[i]==-1:
        stack=[i]
        cnt+=1
        while len(stack)>0:
            tmp=stack.pop()
            union[tmp]=cnt
            for item in tree[tmp]:
                if union[item]==-1:
                    union[item]=cnt
                    stack.append(item)

if cnt==0:
    print(0)
    exit()

cost=[[] for _ in range(cnt+1)]
#print(union)
for i,aa in enumerate(a):
    cost[union[i]].append(aa)

import heapq

del_list=[]
kouho_list=[]

ans=0

#print(cost)
for i in range(cnt+1):
    tmp=min(cost[i])
    ans+=tmp
    del_list.append(tmp)
    kouho_list+=cost[i]
    #print(cost)

#print(cost)
heapq.heapify(kouho_list)
heapq.heapify(del_list)

for _ in range(2*(n-m-1)-(cnt+1)):
    #print(kouho_list,del_list)
    while 1:
        #print(kouho_list,del_list)
        if len(kouho_list)>0:
            tmp=heapq.heappop(kouho_list)
        else:
            print("Impossible")
            exit()
        
        if len(del_list)>0:
            if del_list[0]!=tmp:
                break
            else:
                heapq.heappop(del_list)
        else:
            break
        
    ans+=tmp

print(ans)

