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>

#define F first
#define S second
#define MP make_pair
#define pb push_back
#define all(a) a.begin(), a.end()
#define lcm(a, b) (a)/__gcd((a),(b))*(b)
#define endl '\n'

using namespace std;

typedef long long LL;
typedef pair<int, int> P;
typedef pair<LL, LL> LP;

static const int INF = INT_MAX;
static const LL LINF = LLONG_MAX;
static const int MIN = INT_MIN;
static const LL LMIN = LLONG_MIN;
static const int MOD = 1000000007;
static const int SIZE = 200005;

const int dx[] = {0, -1, 1, 0};
const int dy[] = {-1, 0, 0, 1};

void ufInit(int n);
int Root(int x);
bool Unite(int x, int y);

void dfs(int pos, vector<int>& v);

int root[SIZE];
int sz[SIZE];
int cost[SIZE];

vector<int> graph[SIZE];
bool isvisit[SIZE];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, m;
    cin >> n >> m;
    ufInit(n);

    for(int i = 0; i < n; ++i) {
        cin >> cost[i];

    }

    while(m--) {
        int x, y;
        cin >> x >> y;
        Unite(x, y);
        graph[x].pb(y);
        graph[y].pb(x);

    }

    for(int i = 0; i < n; ++i) {
        Root(i);

    }

    int cnt = 0;
    vector<int> roots;
    LL res = 0;
    set<int> st;
    for(int i = 0; i < n; ++i) {
        if(sz[i] == 1) {
            ++cnt;
            res += cost[i];

        } else {
            roots.pb(root[i]);

        }
        st.insert(root[i]);

    }
    
    if(st.size() == 1) {
        cout << 0 << endl;
        return 0;

    }

    LL unite_cnt = st.size() * 2 - 2;
    unite_cnt -= cnt;
    if(cnt > 2) {
        cout << "Impossible" << endl;
        return 0;

    }

    sort(all(roots));
    roots.erase(unique(all(roots)), roots.end());

    vector<int> unite_cost;
    for(int i = 0; i < roots.size(); ++i) {
        vector<int> tmp;
        dfs(roots[i], tmp);
        if(tmp.size() == 0) {
            continue;

        }
        --unite_cnt;
        sort(all(tmp));
        res += tmp[0];
        for(int j = 1; j < tmp.size(); ++j) {
            unite_cost.pb(tmp[j]);

        }

    }

    if(st.size() == 2) {
        cout << res << endl;
        return 0;

    }
    sort(all(unite_cost));

    for(int i = 0; i < unite_cnt; ++i) {
        res += unite_cost[i];

    }
    cout << res << endl;

    return 0;
}

void ufInit(int n) {
    for(int i = 0; i <= n; ++i) {
        root[i] = i;
        sz[i] = 1;

    }

    return;
}

int Root(int x) {
    if(x == root[x]) {
        return x;

    }
    int rt = Root(root[x]);
    sz[x] = sz[rt];
    root[x] = rt;
    return rt;
}

bool Unite(int x, int y) {
    int rx, ry;
    rx = Root(x);
    ry = Root(y);

    if(rx == ry) {
        return false;

    }
    
    if(sz[rx] > sz[ry]) {
        root[ry] = rx;

    } else {
        root[rx] = ry;

    }
    sz[rx] += sz[ry];
    sz[ry] = sz[rx];

    return true;
}

void dfs(int pos, vector<int>& v) {
    if(isvisit[pos]) {
        return;

    }
    isvisit[pos] = true;
    v.pb(cost[pos]);

    for(int i = 0; i < graph[pos].size(); ++i) {
        dfs(graph[pos][i], v);

    }

    return;
}