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<cstdio>
#include<cstring>
#include<iostream>
#include<map>
#include<queue>
#include<iostream>
#include<cmath>
#include<stack>
using namespace std;
typedef long long LL;

const int maxn = 1e5 + 5;
vector<int> G[maxn];
int low[maxn],dfn[maxn],in[maxn];
bool vis[maxn];
stack<int> st;

LL cost[maxn];
priority_queue<LL, vector<LL>, greater<LL> > que[maxn];
int n,m,ind,cnt;
int st1[maxn];
int head = 1,tail = 0;

void dfs(int x){
	low[x] = dfn[x] = ++ind;
	st.push(x),vis[x] = 1;
	for(int i = 0;i < G[x].size();i++){
		int v = G[x][i];
		if(!dfn[v]){
			dfs(v);
			low[x] = min(low[x],low[v]);
		}
		else if(vis[v]){
			low[x] = min(low[x],dfn[v]);
		}
	}
	if(low[x] == dfn[x]){
		cnt++;
		bool flag = false;
		while(!st.empty()){
			int tot = st.top();
			st.pop();
			vis[tot] = 0;
			in[tot] = 0;
			in[tot] = cnt;
			flag = true;
			que[cnt].push(cost[tot]);
			if(tot == x) break;
		}
		if(flag) st1[++tail] = cnt;
	}
}

int main(){
	LL sum = 0;
	scanf("%d %d",&n,&m);
	for(int i = 0;i < n;i++) scanf("%lld",&cost[i]);
	for(int i = 1;i <= m;i++){
		int x,y;
		scanf("%d %d",&x,&y);
		G[x].push_back(y);
		G[y].push_back(x);
	}
	if(n < 2 * (n - m - 1)){
		printf("Impossible\n");
		return 0;
	}
	else if(n == m + 1){
		printf("0\n");
		return 0;
	}
	for(int i = 0;i < n;i++){
		if(!dfn[i]) dfs(i);
	}
	while(true){
		if(tail == head) break;
		int x = st1[tail];tail--;
		int y = st1[tail];tail--;
		if(!que[x].empty() && !que[y].empty()){
			sum += que[x].top() + que[y].top();
		}
		else{
			printf("Yes\n");
			return 0;
		}
		while(!que[x].empty()){
			LL ans = que[x].top();
			que[x].pop();
			que[y].push(ans);
		}
		st1[++tail] = y;
	}
	printf("%lld\n",sum);
	return 0;
}