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;

const int MX = 100000;

namespace dsu {
	int p[MX];
	
	void init(int n) {
		iota(p, p + n, 0);
	}
	
	int get(int v) {
		if (p[v] != v) p[v] = get(p[v]);
		return p[v];
	}
	
	bool merge(int u, int v) {
		u = get(u);
		v = get(v);
		p[u] = v;
		return u != v;
	}
}

int a[MX];
vector<int> v[MX];

int main() {
	int n, m;
	ignore = scanf("%d %d", &n, &m);
	
	if (m == n - 1) {
		printf("0\n");
		return 0;
	}
	
	for (int i = 0; i < n; i++) ignore = scanf("%d", a + i);
	
	dsu::init(n);
	
	while (m--) {
		int u, v;
		ignore = scanf("%d %d", &u, &v);
		dsu::merge(u, v);
	}
	
	for (int i = 0; i < n; i++) v[dsu::get(i)].push_back(a[i]);
	
	long long ans = 0;
	vector<int> vec;
	int cnt = 0;
	for (int i = 0; i < n; i++)
		if (v[i].empty() == false) {
			sort(v[i].begin(), v[i].end());
			reverse(v[i].begin(), v[i].end());
			ans += v[i].back();
			v[i].pop_back();
			copy(v[i].begin(), v[i].end(), back_inserter(vec));
			cnt++;
		}
	
	if (vec.size() < cnt - 2) {
		printf("Impossible\n");
		return 0;
	}
	
	sort(vec.begin(), vec.end());
	for (int i = 0; i < cnt - 2; i++) ans += vec[i];
	
	printf("%lld\n", ans);
	
	return 0;
}
