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
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**7)
import heapq

n, m = map(int, input().split())
if m == n-1:
  print(0)
  exit()
if n < 2*(n-m-1):
  print("Impossible")
  exit()
A = tuple(map(int, input().split()))
G = [[] for _ in range(n)]
for _ in range(m):
  x, y = map(int, input().split())
  G[x].append(y)
  G[y].append(x)
ans = 0
L = []
seen = [False]*n
def dfs(v):
  global ans
  K = []
  stack = [v]
  while stack:
    v = stack.pop()
    heapq.heappush(K, A[v])
    seen[v] = True
    for nv in G[v]:
      if seen[nv]:
        continue
      stack.append(nv)
  ans += heapq.heappop(K)
  for k in K:
    heapq.heappush(L, k)
for i in range(n):
  if not seen[i]:
    dfs(i)
for _ in range(n-m-2):
  ans += heapq.heappop(L)
print(ans)