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;
typedef long long ll;

vector<int> G[100000];
ll c[100000];
int vis[100000];
int id[100000];
priority_queue<int, vector<int>, greater<int>> q[100000];

void dfs(int v, int p, int i){
	id[v] = i;
	q[i].push(c[v]);
	vis[v] = 1;
	for(auto c : G[v]){
		if(c == p) continue;
		dfs(c, v, i);
	}
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
#ifdef LOCAL
	std::ifstream in("in");
	std::cin.rdbuf(in.rdbuf());
#endif

	int N, M;
	cin >> N >> M;

	if(M == N - 1){
		cout << 0 << endl;
		return 0;
	}

	for(int i = 0; i < N; i++){
		cin >> c[i];
	}
	for(int i = 0; i < M; i++){
		int x, y;
		cin >> x >> y;
		G[x].push_back(y);
		G[y].push_back(x);
	}

	int T = 0;
	for(int i = 0; i < N; i++){
		if(!vis[i]){
			dfs(i, -1, T);
			T++;
		}
	}

	ll ans = 0;
	priority_queue<int, vector<int>, greater<int>> Q;
	for(int i = 0; i < T; i++){
		ans += q[i].top();
		q[i].pop();
		while(q[i].size()){
			Q.push(q[i].top());
			q[i].pop();
		}
	}

	if(Q.size() < T - 2){
		cout << "Impossible" << endl;
		return 0;
	}

	for(int i = 0; i < T - 2; i++){
		ans += Q.top();
		Q.pop();
	}
	cout << ans << endl;
}
