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 = [int(i) for i in input().split()]
xy = [[int(i) for i in input().split()] for _ in range(M)]

class UnionFind() :
    def __init__(self, N) :
        self.parent = [-1] * N
        
    def root(self, x) :
        while self.parent[x] >= 0:
            x = self.parent[x]
        return x

    def unite(self, x, y) :
        rx = self.root(x)
        ry = self.root(y)

        if rx != ry :
            if self.parent[rx] > self.parent[ry] :
                self.parent[rx] = ry
            else :
                if self.parent[rx] == self.parent[ry] :
                    self.parent[rx] -= 1
                self.parent[ry] = rx
    
    def same(self, x, y) :
        return self.root(x) == self.root(y)
        
    def size(self, x) :
        return -self.parent[self.root(x)]
        
UF = UnionFind(N)
for x, y in xy :
    UF.unite(x, y)

component = 0
for p in UF.parent :
    if p < 0 :
        component += 1
        
if N < 2 * (component - 1) :
    print('Impossible')
elif component == 1 :
    print(0)
else :
    ret = 0
    b = [(a[i], UF.root(i)) for i in range(N)]
    b.sort()
    visited = [False] * N
    c = component - 2
    for v, k in b :
        if not visited[k] :
            ret += v
            visited[k] = True
        elif c > 0 :
            ret += v
            c -= 1
            
    print(ret)