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: Accepted)

#include<iostream>
#include<vector>
#include<set>
#include<queue>
using namespace std;

struct UF{
    vector<int> p;
    int n;

    UF(int siz){
        n = siz;
        p.resize(n, 0);
        for(int i = 0; i < n; i++)  p[i] = i;
    }

    int parent(int x){
        if(p[x] != x)   p[x] = parent(p[x]);
        return p[x];
    }

    bool same(int x, int y){
        return parent(x) == parent(y);
    }
    
    void unite(int x, int y){
        x = parent(x), y = parent(y);
        p[x] = y;
    }
};

typedef long long ll;

int main(){
    int n, m;
    cin >> n >> m;
    if((n-1-m)*2 > n){
        cout << "Impossible" << endl;
        return 0;
    }else if(n-1 == m){
        cout << 0 << endl;
        return 0;
    }
    int a[n];
    for(int i = 0; i < n; i++)  cin >> a[i];
    UF uf(n);
    for(int i = 0; i < m; i++){
        int x, y;
        cin >> x >> y;
        uf.unite(x, y);
    }
    multiset<int> s[n];
    for(int i = 0; i < n; i++)  s[uf.parent(i)].insert(a[i]);
    priority_queue<int> pq;
    ll ans = 0;
    int res = (n-1-m)*2;
    for(int i = 0; i < n; i++){
        if(s[i].empty())    continue;
        ans += *s[i].begin();
        res--;
        for(auto it = ++s[i].begin(); it != s[i].end(); it++)   pq.push({-(*it)});
    }
    while(res-- > 0){
        ans += -pq.top();   pq.pop();
    }
    cout << ans << endl;
    return 0;
}