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, m;
int a[100005];
struct Union_Find_Set {
	int fa[100005];
	void Init() {
		for (int i = 1; i <= n; i++) {
			fa[i] = i;
		}
	}
	int Find(int x) {
		if (fa[x] != x) fa[x] = Find(fa[x]);
		return fa[x];
	}
	void Union(int x, int y) {
		fa[Find(x)] = Find(y);
	}
}ufs;
vector<int>vec[100005], v;
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
	}
	int rem = (n - 1 - m) * 2;
	if (n < rem) {
		printf("Impossible\n");
		return 0;
	}
	if (rem == 0) {
		printf("0\n");
		return 0;
	}
	ufs.Init();
	for (int x, y, i = 1; i <= m; i++) {
		scanf("%d%d", &x, &y);
		x++;
		y++;
		ufs.Union(x, y);
	}
	for (int i = 1; i <= n; i++) {
		vec[ufs.Find(i)].push_back(a[i]);
	}
	long long ans = 0;
	for (int i = 1; i <= n; i++) {
		if (ufs.Find(i) == i) {
			sort(vec[i].begin(), vec[i].end());
			ans += vec[i][0];
			rem--;
			for (int j = 1; j < int(vec[i].size()); j++) {
				v.push_back(vec[i][j]);
			}
		}
	}
	sort(v.begin(), v.end());
	for (int i = 0; i < rem; i++) {
		ans += v[i];
	}
	printf("%lld\n", ans);
	return 0;
}