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;
long long A[100010]; 
bool seen[100010]; 
vector<int> E[100010]; 
vector<int> C[100010]; 
void dfs(int now, int parent, int &id){
	seen[now] = true; 
	C[id].push_back(A[now]); 
	for(int i = 0; i < E[now].size(); i++){
		if(E[now][i] == parent) continue; 
		dfs(E[now][i], now, id); 
	}
}
int main(){
	cin >> N >> M; 
	for(int i = 0; i < N; i++){
		cin >> A[i]; 
	}
	for(int i = 0; i < M; i++){
		int x, y; cin >> x >> y; 
		E[x].push_back(y); E[y].push_back(x); 
	}
	int id = 0; 
	for(int i = 0; i < N; i++){
		if(seen[i]) continue;
		id ++; 
		dfs(i, -1, id);  
	}
	long long ans = 0; 
	priority_queue<long long, vector<long long>, greater<long long>> que;  
	for(int i = 1; i <= id; i++){
		sort(C[i].begin(), C[i].end()); 
		ans += C[i][0]; 
		for(int j = 1; j < C[i].size(); j++) que.push(C[i][j]); 		
	}
	if(N > 2 * M + 2){
		cout << "Impossible" << endl; return 0; 
	}
	if(N == M + 1){
		cout << 0 << endl; return 0; 
	}
	for(int i = 0; i < 2 * (N-M-1) - id; i++){
		long long a = que.top(); 
		ans += a; que.pop(); 
	}
	cout << ans << endl; 
}