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
#include <bits/stdc++.h>
using namespace std;
int n,k,fa[100005],size[100005];
pair<int,int> a[100005];
int z(int x){
	if (x == fa[x]) return x;
	fa[x] = z(fa[x]);
	return fa[x];
}
void bin(int x,int y){
	x = z(x);
	y = z(y);
	fa[x] = y;
	size[y] += size[x];
}
int main(){
	scanf("%d",&n);
	scanf("%d",&k);
	for (int i = 0; i < n; i++)
		fa[i] = i;
	for (int i = 0; i < n; i++)
		size[i] = 1;
	long long ans = 0;
	for (int i = 0; i < n; i++){
		scanf("%d",&a[i].first);
		ans += a[i].first;
	}
	for (int i = 0; i < n; i++)
		a[i].second = i;
	sort(a, a + n);
	for (int i = 1; i <= k; i++){
		int x,y;
		scanf("%d%d",&x,&y);
		bin(x,y);
	}
	if (n - k == 1){
		printf("0\n");
		return 0;
	}
	int s = n - (n - k - 1) * 2;
	if (s < 0){
		printf("Impossible\n");
		return 0;
	}
	for (int i = n - 1; (i >= 0) && (s > 0); i--)
		if (size[z(a[i].second)] > 1){
			size[z(a[i].second)]--;
			s--;
			ans -= a[i].first;
		}
	printf("%lld\n",ans);
}