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 <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

typedef long long ll;
const int INF = 1e9;
const int MAXN = 1e5+10;

int n, m;
vector<int> G[MAXN];
ll A[MAXN];
bool chk[MAXN];
vector<int> gr[MAXN];
int gcnt=0;

void dfs(int cur) {
	gr[gcnt].push_back(A[cur]);
	chk[cur] = true;
	for (auto nxt: G[cur]) {
		if(chk[nxt]) continue;
		dfs(nxt);
	}
}

int main() {
	scanf("%d %d", &n, &m);
	
	for (int i = 0; i < n; i++)
		scanf("%d", &A[i]);

	for (int i = 0; i < m; i++) {
		int u, v;
		scanf("%d %d", &u, &v);
		G[u].push_back(v);
		G[v].push_back(u);
	}

	for (int i = 0; i < n; i++) {
		if(chk[i]) continue;
		
		dfs(i);
		sort(gr[gcnt].begin(), gr[gcnt].end());
		gcnt++;
	}

	if (gcnt == 1) {
		puts("0");
		return 0;
	}

	// 필요한 간선의 개수 = gr.size()-1
	if ((gcnt - 1)*2 > n) {
		puts("Impossible");
		return 0;
	}

	
	vector<int> cand;
	ll ans = 0;
	for (int i = 0; i < gcnt; i++) {
		ans += gr[i][0];
		for (int j = 1; j < gr[i].size(); j++)
			cand.push_back(gr[i][j]);
	}

	sort(cand.begin(), cand.end());
	int rem = gcnt - 2;
	while(rem--)
		ans += cand[rem];
	
	printf("%lld", ans);
	return 0;
}