Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

7 5
1 2 3 4 5 6 7
3 0
4 0
1 2
1 3
5 6

Sample Output 1

7

If we connect vertices 0 and 5, the graph becomes connected, for the cost of 1 + 6 = 7 dollars.

Sample Input 2

5 0
3 1 4 1 5

Sample Output 2

Impossible

We can't make the graph connected.

Sample Input 3

1 0
5

Sample Output 3

0

The graph is already connected, so we do not need to add any edges.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
int N,M;
int T[100005];
pair<int,int> A[100005];
priority_queue<pair<int,int>,vector<pair<int,int> >,greater<pair<int,int> > > H;
int fi(int nod){
    if(!T[nod]){
        return nod;
    }
    return T[nod] = fi(T[nod]);
}
void u(int x,int y){
    T[x] = y;
}
int main(){
    cin >> N >> M;
    for(int i = 1;i <= N;i++){
        cin >> A[i].first;
        A[i].second = i;
    }
    for(int i = 1;i <= M;i++){
        int x,y;
        cin >> x >> y;
        x++;
        y++;
        u(fi(x),fi(y));
    }
    sort(A + 1,A + 1 + N);
    int nr = M;
    long long rez = 0;
    for(int i = 1;i <= N && nr < N - 1;i++){
        if(H.empty() || fi(H.top().second) == fi(A[i].second)){
            H.push(A[i]);
        }
        else{
            rez += H.top().first + A[i].first;
            u(fi(H.top().second),A[i].second);
            H.pop();
            nr++;
        }
    }
    if(nr < N - 1){
        cout << "Impossible";
    }
    else{
        cout << rez;
    }
    return 0;
}