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<bits/stdc++.h>
using namespace std;
vector<long long>adj[100005];
long long a[100005];
bool visited[100005];
void dfs(int x)
{
    
    if(visited[x])return ;
    visited[x]=true;
    for(int i=0;i<adj[x].size();i++)
    {
        if(!visited[i])dfs(adj[x][i]);
    }
}
int main()
{


  int n,m;
  scanf("%d%d",&n,&m);
  for(int i=0;i<n;i++)
    scanf("%lld",&a[i]);
    for(int i=0;i<m;i++)
    {
        int x,y;
        cin>>x>>y;
        adj[x].push_back(y);
        adj[y].push_back(x);
    }
    memset(visited,false,sizeof(visited));
    for(int i=0;i<n;i++)
    {
        if(!visited[i])
            dfs(i);
    }
}