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
def main():
    import sys,heapq
    input = sys.stdin.readline

    n,m = map(int,input().split())
    a = tuple(map(int,input().split()))


    #Union Find
    par = [-1]*n
    #xの根を求める
    def find(x):
        if par[x] < 0:
            return x
        else:
            par[x] = find(par[x])
            return par[x]
    #xとyの属する集合の併合
    def unite(x,y):
        x = find(x)
        y = find(y)

        if x == y:
            return False
        else:
            #sizeの大きいほうがx
            if par[x] > par[y]:
                x,y = y,x
            par[x] += par[y]
            par[y] = x
            return True

    #xとyが同じ集合に属するかの判定
    def same(x,y):
        return find(x) == find(y)

    #xが属する集合の個数
    def size(x):
        return -par[find(x)]


    for _ in [0]*m:
        x,y = map(int,input().split())
        unite(x,y)
    
    if m < (n-1)//2:
        print('Impossible')
        return
    
    h = [[] for _ in [0]*n]
    for i,e in enumerate(a):
        heapq.heappush(h[find(i)],e)
    
    root = []
    for i,e in enumerate(par):
        if e < 0:
            root.append(i)
    
    if len(root) == 1:
        print(0)
        return

    must = 0
    opt = []
    for i in root:
        must += heapq.heappop(h[i])
        for e in h[i]:
            opt.append(e)
    opt.sort()

    print(must+sum(opt[:len(root)-2]))



if __name__ =='__main__':
    main()