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 "bits/stdc++.h"
using namespace std;
#define ll long long int
#define rep(i,n) for( int i = 0; i < n; i++ )
#define rrep(i,n) for( int i = n; i >= 0; i-- )
#define REP(i,s,t) for( int i = s; i <= t; i++ )
#define RREP(i,s,t) for( int i = s; i >= t; i-- )
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define INF 2000000000
#define mod 1000000007
#define INF2 1000000000000000000
#define int long long
typedef pair<int, int> P;
priority_queue<P, vector<P>, greater<P> > que[100010];
vector<int> G[100010];
bool used[100010];
bool used2[100010];
P a[100010];
int size = 0;
void dfs(int x) {
    used[x] = true;
    que[size].push({a[x].first, x});
    for(int next: G[x]) {
        if (!used[next]) dfs(next);
    }
}

signed main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N, M;
    cin >> N >> M;
    rep(i, N) {
        cin >> a[i].first;
        a[i].second = i;
    }
    if (M == N - 1) {
        cout << 0 << endl;
        return 0;
    }
    rep(i, M) {
        int x, y;
        cin >> x >> y;
        G[x].push_back(y);
        G[y].push_back(x);
    }
    rep(i, N) if (!used[i]) {dfs(i); size++;}
    //cout << size << endl;
    int ans = 0;
    rep(i, size) {
        ans += que[i].top().first;
        used2[que[i].top().second] = true;
    }
    int cnt = 0;
    sort(a, a + N);
    int i = 0;
    while(i < N && cnt < size - 2) {
        //cout << a[i].second << " " << !used2[a[i].second] << endl;
        if (!used2[a[i].second]) {ans += a[i].first; cnt++;}
        i++;
    }
    if (cnt != size - 2) {
        cout << "Impossible" << endl;
    } else {
        cout << ans << endl;
    }



    return 0;
}