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)

// May this submission get accepted!

#pragma GCC optimize ("O3")
#pragma GCC target ("tune=native")
#pragma GCC target ("avx")

#include <bits/stdc++.h>

// 汎用マクロ
#define ALL_OF(x) (x).begin(), (x).end()
#define REP(i,n) for (long long i=0, i##_len=(n); i<i##_len; i++)
#define RANGE(i,is,ie) for (long long i=(is), i##_end=(ie); i<=i##_end; i++)
#define DSRNG(i,is,ie) for (long long i=(is), i##_end=(ie); i>=i##_end; i--)
#define UNIQUE(v) { sort((v).begin(), (v).end()); (v).erase(unique((v).begin(), (v).end()), (v).end()); }
template<class T> bool chmax(T &a, const T &b) {if (a < b) {a = b; return true;} return false; }
template<class T> bool chmin(T &a, const T &b) {if (a > b) {a = b; return true;} return false; }
#define INF 0x7FFFFFFF
#define LINF 0x7FFFFFFFFFFFFFFFLL
#define Yes(q) ((q) ? "Yes" : "No")
#define YES(q) ((q) ? "YES" : "NO")
#define Possible(q) ((q) ? "Possible" : "Impossible")
#define POSSIBLE(q) ((q) ? "POSSIBLE" : "IMPOSSIBLE")
#define DUMP(q) cerr << "[DEBUG] " #q ": " << (q) << " at " __FILE__ ":" << __LINE__ << endl
#define DUMPALL(q) { cerr << "[DEBUG] " #q ": ["; REP(i, (q).size()) { cerr << (q)[i] << (i == i_len-1 ? "" : ", "); } cerr << "] at " __FILE__ ":" << __LINE__ << endl; }
template<class T> T gcd(const T &a, const T &b) { return a < b ? gcd(b, a) : b ? gcd(b, a % b) : a; }
template<class T> T lcm(const T &a, const T &b) { return a / gcd(a, b) * b; }

// gcc拡張マクロ
#define popcount __builtin_popcount
#define popcountll __builtin_popcountll

// エイリアス
#define DANCE_ long
#define ROBOT_ unsigned
#define HUMAN_ signed
#define CHOKUDAI_ const
using  ll = DANCE_ HUMAN_ DANCE_;
using ull = DANCE_ ROBOT_ DANCE_;
using cll = DANCE_ DANCE_ CHOKUDAI_;
using  ld = long double;
using namespace std;

// モジュール
// Union-Find モジュール
struct union_find {
    vector<ull> par;
    vector<ull> num;
    union_find(ull n) : par(n, 0), num(n, 1) { iota(ALL_OF(par), 0); }
    ull root(ull i) { return par[i] == i ? i : (par[i] = root(par[i])); }
    bool same(ull i, ull j) { return root(i) == root(j); }
    void unite(ull i, ull j) {
        ull ri = root(i);
        ull rj = root(j);
        if (ri != rj) {
            if (num[ri] > num[rj]) swap(ri, rj);
            par[ri] = rj;
            num[rj] += num[ri];
        }
    }
};

// 処理内容
int main() {
    
    ll n, m; cin >> n >> m;
    vector<ll> a(n), x(m), y(m);
    // already 0-indexed
    REP(i, n) cin >> a[i];
    REP(i, m) cin >> x[i] >> y[i];

    ll reme = n - m - 1;
    if (reme * 2 > n) {
        cout << "Impossible\n";
        return 0;
    }
    
    vector<ll> rank(n);
    iota(ALL_OF(rank), 0LL);
    sort(ALL_OF(rank), [&](ll i, ll j) {
        return a[i] < a[j];
    });

    union_find uf(n);
    REP(i, m) {
        uf.unite(x[i], y[i]);
    }

    vector<bool> used(n, false);
    ll ans = 0;
    REP(_from, n) {
        if (reme <= 0) break;
        ll from = rank[_from];
        if (used[from]) continue;
        RANGE(_to, _from+1, n-1) {
            ll to = rank[_to];
            if (used[to]) continue;
            if (!uf.same(from, to)) {
                ans += a[from] + a[to];
                used[from] = used[to] = true;
                reme--;
                break;
            }
        }
    }
    cout << ans << endl;

}