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;

typedef long long   signed int LL;
typedef long long unsigned int LU;

#define incID(i, l, r) for(int i = (l)    ; i <  (r); i++)
#define incII(i, l, r) for(int i = (l)    ; i <= (r); i++)
#define decID(i, l, r) for(int i = (r) - 1; i >= (l); i--)
#define decII(i, l, r) for(int i = (r)    ; i >= (l); i--)
#define  inc(i, n) incID(i, 0, n)
#define inc1(i, n) incII(i, 1, n)
#define  dec(i, n) decID(i, 0, n)
#define dec1(i, n) decII(i, 1, n)

#define inII(v, l, r) ((l) <= (v) && (v) <= (r))
#define inID(v, l, r) ((l) <= (v) && (v) <  (r))

#define PB push_back
#define EB emplace_back
#define MP make_pair
#define FI first
#define SE second
#define PQ priority_queue

#define  ALL(v)  v.begin(),  v.end()
#define RALL(v) v.rbegin(), v.rend()
#define  FOR(it, v) for(auto it =  v.begin(); it !=  v.end(); ++it)
#define RFOR(it, v) for(auto it = v.rbegin(); it != v.rend(); ++it)

template<typename T> bool   setmin(T & a, T b) { if(b <  a) { a = b; return true; } else { return false; } }
template<typename T> bool   setmax(T & a, T b) { if(b >  a) { a = b; return true; } else { return false; } }
template<typename T> bool setmineq(T & a, T b) { if(b <= a) { a = b; return true; } else { return false; } }
template<typename T> bool setmaxeq(T & a, T b) { if(b >= a) { a = b; return true; } else { return false; } }
template<typename T> T gcd(T a, T b) { return (b == 0 ? a : gcd(b, a % b)); }
template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; }

// ---- ----

struct UnionFind {
	int s, num; // [0, s) の s 個の要素がある。num は非空集合の個数。
	vector<vector<int>> v;
	vector<int> find;
	
	UnionFind() { s = 0; }
	UnionFind(int s) { init(s); }
	void init(int arg_s) {
		num = s = arg_s;
		v.clear();
		find.clear();
		inc(i, s) { v.PB({i}); find.PB(i); }
	}
	bool unite(int x, int y) {
		assert(inID(x, 0, s) && inID(y, 0, s));
		if(find[x] == find[y]) { return false; }
		num--;
		if(v[find[x]].size() > v[find[y]].size()) { swap(x, y); }
		int fx = find[x];
		for(auto && e: v[fx]) { v[find[y]].PB(e); find[e] = find[y]; }
		v[fx].clear();
		return true;
	}
};

// ----

LL n, m, a[100000];
vector<LL> v[100000], w;

int main() {
	cin >> n >> m;
	inc(i, n) { cin >> a[i]; }
	UnionFind uf(n);
	inc(i, m) {
		int x, y;
		cin >> x >> y;
		uf.unite(x, y);
	}
	
	inc(i, n) { v[uf.find[i]].PB(a[i]); }
	inc(i, n) { sort(ALL(v[i])); }
	
	if(m == n - 1) { cout << 0 << endl; return 0; }
	
	LL ans = 0, c = 0;
	inc(i, n) {
		inc(j, v[i].size()) {
			if(j == 0) { ans += v[i][j]; c++; } else { w.PB(v[i][j]); }
		}
	}
	
	sort(ALL(w));
	bool f = false;
	inc(i, w.size()) {
		if(c == 2 * (n - 1 - m)) { f = true; break; }
		ans += w[i];
	}
	
	if(! f) { cout << "Impossible" << endl; }
	else { cout << ans << endl; }
	
	return 0;
}