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
from collections import defaultdict
from heapq import heappop, heappush
n,m = map(int, input().split())
a = list(map(int, input().split()))

uf = [-1]*n
def root(x):
    if uf[x] < 0:
        return x
    uf[x] = root(uf[x])
    return uf[x]
def unite(x,y):
    rx, ry = root(x), root(y)
    if rx == ry:
        return False
    if uf[rx] > uf[ry]:
        rx, ry = ry, rx
    uf[rx] += uf[ry]
    uf[ry] = rx
    return True

for i in range(m):
    x,y = map(int, input().split())
    unite(x,y)
if m == n-1:
    print(0)
elif (n-m-1)*2 > n:
    print("Impossible")
else:
    d = defaultdict(list)
    for i in range(n):
        heappush(d[root(i)], a[i])
    cnt = (n-m-1)*2
    ans = 0
    heap = []
    for value in d.values():
        ans += heappop(value)
        cnt -= 1
        while value:
            heappush(heap, heappop(value))
    for _ in range(cnt):
        ans += heappop(heap)
    print(ans)