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
from collections import deque, defaultdict
import heapq

sr = lambda: sys.stdin.readline().rstrip()
ir = lambda: int(sr())
lr = lambda: list(map(int, sr().split()))

N, M = lr()
A = lr()

# Aでxの根を求める
def find(A,x):
    p = A[x]
    if p == x:
        return x
    a = find(A,p)
    A[x] = a
    return a

# Aでxとyの属する集合の併合
def union(A, x, y):
    if find(A,x) > find(A,y):
        bx, by = find(A,y), find(A,x)
    else:
        bx, by = find(A,x), find(A,y)
    # 根をbxに統一
    A[by] = bx

edges = [[] for _ in range(N)]
V = [i for i in range(N)]
for _ in range(M):
    x, y = lr()
    if find(V, x) != find(V, y):
        union(V, x, y)
    edges[x].append(y)
    edges[y].append(x)

for i in range(N):
    V[i] = find(V, i)

if M == N - 1:
    print(0); exit()
elif N < 2 * (N-M-1):
    print('Impossible'); exit()

root_point = defaultdict(list)
for i in range(N):
    r = V[i]
    heapq.heappush(root_point[r], A[i])

#各グループの最小値以外で使うコストの数
answer = 0
candidates = []
for p_list in root_point:
    answer += heapq.heappop(root_point[p_list])
    candidates.extend(root_point[p_list])

remain = 2 * (N-M-1) - len(set(V))
candidates.sort()
answer += sum(candidates[:remain])
print(answer)
