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 <algorithm>
#include <cstdio>
#include <vector>
#include <set>
int fa[100005], arr[100005];
std::vector<int> vec[100005];
std::multiset<int> se;
int getfa(int u) { return u == fa[u] ? u : fa[u] = getfa(fa[u]); }
int main()
{
	// freopen("K.in", "r", stdin);
	int n, m;
	scanf("%d%d", &n, &m);
	if (m == n - 1)
	{
		puts("0");
		return 0;
	}
	for (int i = 0; i < n; i++)
	{
		scanf("%d", arr + i);
		se.insert(arr[i]);
		fa[i] = i;
	}
	for (int i = 0; i < m; i++)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		fa[getfa(u)] = getfa(v);
	}
	for (int i = 0; i < n; i++)
		vec[getfa(i)].push_back(arr[i]);
	long long ans = 0;
	for (int i = 0; i < n; i++)
	{
		if (vec[i].empty())
			continue;
		int x = *std::min_element(vec[i].begin(), vec[i].end());
		ans += x;
		se.erase(se.find(x));
	}
	for (int i = 0; i < n - m - 2; i++)
	{
		if (se.empty())
		{
			puts("Impossible");
			return 0;
		}
		ans += *se.begin();
		se.erase(se.begin());
	}
	printf("%lld\n", ans);
	return 0;
}
