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 heapq import heapify, heappop, heappush


def solve():
    N, M = map(int, input().split())
    a = list(map(int, input().split()))

    if M == N - 1:
        ans = 0
    elif 2 * (N - M - 1) > N:
        ans = 'Impossible'
    else:
        g2i = [[i] for i in range(N)]
        i2g = list(range(N))

        def merge(ia, ib):
            # print(g2i, i2g, ia, ib)
            if len(g2i[i2g[ia]]) < len(g2i[i2g[ib]]): ia, ib = ib, ia
            ga, gb = i2g[ia], i2g[ib]
            for j in g2i[gb]: i2g[j] = ga
            g2i[ga].extend(g2i[gb])
            g2i[gb] = []

        for _ in range(M):
            tx, ty = map(int, input().split())
            merge(tx, ty)

        ans = 0
        need = 2 * (N - M - 1)
        rest = []
        for gi in range(len(g2i)):
            t = []
            for i in g2i[gi]:
                t.append(a[i])
            if not t: continue
            heapify(t)
            ans += heappop(t)
            need -= 1
            rest += t
        ans += sum(sorted(rest)[:need])

    print(ans)


if __name__ == '__main__':
    solve()
