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>
// #pragma GCC optimize ("O3")
// #pragma GCC target ("sse4")
 
using namespace std;
 
#define li long long int
#define ld long double
const int MX1=1e5+5;
const int MX2=1e6+5;
const li inf=1e18;
const li mod=1e9+7;
 
li power(li a,li b){li res=1;while(b){if(b%2)res=(res*a)%mod;
a=(a*a)%mod;b/=2;}return res;}
 
li mmi(li a){return power(a,mod-2);}
 
li cntbit(li n){li cnt=0;while(n){n&=(n-1);cnt++;}return cnt;}
 
int n,m;
int a[MX1];
int vis[MX1];
int use[MX1];
int mark[MX1];
int arr[MX1];
int sz[MX1];
int comp;
vector<int> g[MX1];
vector<pair<int,int> > v;

int root(int i){
    while(i!=arr[i]){
        i=arr[arr[i]];
    }
    return i;
}

void unionof(int a,int b){
    int roota=root(a);
    int rootb=root(b);
    if(sz[roota]<sz[rootb]) swap(roota,rootb);
    sz[roota]+=sz[rootb];
    arr[rootb]=roota;
}
 
void dfs(int u){
    vis[u]=1;
    mark[u]=comp;
    for(auto v:g[u]){
        if(!vis[v]){
            dfs(v);
        }
    }
}

void solve(){
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        v.push_back(make_pair(a[i],i));
        arr[i]=i;
        sz[i]=1;
    }
    sort(v.begin(),v.end());
    for(int i=0;i<m;i++){
        int x,y;
        cin>>x>>y;
        ++x;
        ++y;
        unionof(x,y);
        g[x].push_back(y);
        g[y].push_back(x);
    }
    for(int i=1;i<=n;i++){
        if(!vis[i]){
            ++comp;
            dfs(i);
        }
    }
    int cnt=1;
    int j=0;
    li ans=0;
    for(int i=0;i<n;){
        j=max(j,i);
        if(use[v[i].second]){
            i++;
            continue;
        }
        while(j<n){
            if(mark[v[i].second]!=mark[v[j].second] and !use[v[i].second] and !use[v[j].second]){
                int roota=root(v[i].second);
                int rootb=root(v[j].second);
                if(roota==rootb){
                    j++;
                    continue;
                }
                if(sz[roota]==1 and sz[rootb]==1 and cnt+1!=comp){
                    j++;
                    continue;
                }
                ans+=0LL+v[i].first+v[j].first;
                use[v[i].second]=use[v[j].second]=1;
                unionof(v[i].second,v[j].second);
                ++cnt;
                j++;
                break;
            }
            j++;
        }
        if(cnt==comp) break;
        i++;
    }
    if(cnt!=comp){
        cout<<"Impossible";
    }else{
        cout<<ans;
    }
}
 
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t=1;
    //cin>>t;
    while(t--){
        solve();
    }
    return 0;
}