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

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = 10 ** 19
MOD = 10 ** 9 + 7
EPS = 10 ** -10

class UnionFind:
    """ Union-Find木 """

    def __init__(self, n):
        self.n = n
        self.par = [i for i in range(n)]
        self.rank = [0] * n
        self.size = [1] * n
        self.tree = [True] * n
        self.grpcnt = n

    def find(self, x):
        """ 根の検索(グループ番号の取得) """
        t = []
        while self.par[x] != x:
            t.append(x)
            x = self.par[x]
        for i in t:
            self.par[i] = x
        return self.par[x]

    def union(self, x, y):
        """ 併合 """
        x = self.find(x)
        y = self.find(y)

        if x == y:
            self.tree[x] = False
            return
        if not self.tree[x] or not self.tree[y]:
            self.tree[x] = self.tree[y] = False

        self.grpcnt -= 1
        if self.rank[x] < self.rank[y]:
            self.par[x] = y
            self.size[y] += self.size[x]
        else:
            self.par[y] = x
            self.size[x] += self.size[y]
            if self.rank[x] == self.rank[y]:
                self.rank[x] += 1

    def is_same(self, x, y):
        """ 同じ集合に属するか判定 """
        return self.find(x) == self.find(y)

    def get_size(self, x=None):
        if x is not None:
            """ あるノードの属する集合のノード数 """
            return self.size[self.find(x)]
        else:
            """ 集合の数 """
            return self.grpcnt
    
    def is_tree(self, x):
        """ 木かどうかの判定 """
        return self.tree[self.find(x)]

N, M = MAP()
A = LIST()
uf = UnionFind(N)
for i in range(M):
    a, b = MAP()
    uf.union(a, b)

L = uf.get_size()
K = (L-1) * 2 
if K > N:
    print('Impossible')
    exit()
elif K == 0:
    print(0)
    exit()

adjli = [[] for i in range(N)]
for i in range(N):
    root = uf.find(i)
    adjli[root].append(A[i])

ans = 0
B = []
for i in range(N):
    if adjli[i]:
        adjli[i].sort(reverse=1)
        ans += adjli[i].pop()
        B += adjli[i]
        K -= 1

B.sort()
ans += sum(B[:K])
print(ans)
