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()))
XY = [tuple(map(int,input().split())) for i in range(M)]

if M==N-1:
    print(0)
    exit()

class UnionFind:
    def __init__(self,N):
        self.parent = [i for i in range(N)]
        self._size = [1] * N
        self.count = 0
    def root(self,a):
        if self.parent[a] == a:
            return a
        else:
            self.parent[a] = self.root(self.parent[a])
            return self.parent[a]
    def is_same(self,a,b):
        return self.root(a) == self.root(b)
    def unite(self,a,b):
        ra = self.root(a)
        rb = self.root(b)
        if ra == rb: return
        if self._size[ra] < self._size[rb]: ra,rb = rb,ra
        self._size[ra] += self._size[rb]
        self.parent[rb] = ra
        self.count += 1
    def size(self,a):
        return self._size[self.root(a)]

uf = UnionFind(N)
for x,y in XY:
    if uf.is_same(x,y): continue
    uf.unite(x,y)
for i in range(N):
    uf.root(i)

from collections import defaultdict
dic = defaultdict(lambda: set())
for i in range(N):
    dic[uf.root(i)].add(i)

need = (len(dic)-1)*2
if need > N:
    print('Impossible')
    exit()

ans = used = 0
import heapq
hq = []
heapq.heapify(hq)
for r,vs in dic.items():
    mini = -1
    mina = 10**9+1
    for i in vs:
        if A[i] < mina:
            mina = A[i]
            mini = i
    ans += mina
    used += 1
    for i in vs:
        if i==mini: continue
        heapq.heappush(hq, A[i])

for _ in range(need - used):
    ans += heapq.heappop(hq)
print(ans)