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;
typedef long long ll;
typedef pair<int,int> pp;
typedef pair<ll,ll> pll;
void read(int& x){ scanf("%d",&x); }
void read(ll& x){ scanf("%lld",&x); }
template<typename T,typename... Args>
void read(T& a,Args&... b){ read(a); read(b...); }
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define eb emplace_back
#define x first
#define y second
int n, m;
int par[100010];
int r(int x){ return (x==par[x])?x:(par[x]=r(par[x])); }
priority_queue<int> pq[100010];
bool join(int a, int b){
	a=r(a); b=r(b);
	if(a==b) return 0;
	par[a]=b;
	if(pq[a].size()>pq[b].size()) swap(pq[a], pq[b]);
	while(pq[a].size()){
		pq[b].push(pq[a].top()); pq[a].pop();
	}
	return 1;
}
int pc[100010];
inline void dame(){ puts("Impossible"); exit(0); }
int main()
{
	read(n, m);
	for(int i=0; i<n; ++i){
		par[i]=i;
		int x; read(x);
		pq[i].push(-x);
	}
	for(int i=1; i<=m; ++i){
		int a, b; read(a, b); join(a, b);
	}
	priority_queue<int> q;
	ll ans = 0;
	int c=0, o=0;
	for(int i=0; i<n; ++i) if(par[i] == i){
		ans += pq[i].top(); pq[i].pop();
		if(pq[i].empty()) ++o;
		++c;
		while(pq[i].size()){
			q.push(pq[i].top());
			pq[i].pop();
		}
	}
	if(c==1){
		puts("0"); return 0;
	}
	for(int i=0; i<c-2; ++i){
		if(q.empty()){
			puts("Impossible");
			return 0;
		}
		ans += q.top(); q.pop();
	}
	printf("%lld\n", -ans);
	return 0;
}