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.

Examples

Input

7 5
1 2 3 4 5 6 7
3 0
4 0
1 2
1 3
5 6


Output

7


Input

5 0
3 1 4 1 5


Output

Impossible


Input

1 0
5


Output

0
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long int ll;

int n,m;
ll a[100100];
vector<int> G[100100];
int used[100100];
vector<ll> ve;

void dfs(int s){
    used[s]=1;
    ve.push_back(a[s]);
    for(int t:G[s]){
        if(!used[t]){
            dfs(t);
        }
    }
    return;
}

int main(){
    cin >> n >> m;
    for(int i=0;i<n;i++){
        cin >> a[i];
    }
    for(int i=0;i<m;i++){
        int x,y; cin >> x >> y;
        G[x].push_back(y);
        G[y].push_back(x);
    }
    if(n-1==m){
        cout << 0 << endl;
        return 0;
    }
    if(n<2*(n-m-1)){
        cout << "Impossible" << endl;
        return 0;
    }
    vector<ll> v;
    ll ans=0; int cnt=0;
    for(int i=0;i<n;i++){
        if(!used[i]){
            ve.clear();
            dfs(i);
            sort(ve.begin(),ve.end());
            ans+=ve[0];
            cnt++;
            for(int j=1;j<ve.size();j++){
                v.push_back(ve[j]);
            }
        }
    }
    sort(v.begin(),v.end());
    for(int i=0;i<v.size()&&cnt<2*(n-1-m);i++,cnt++){
        ans+=v[i];
    }
    cout << ans << endl;
}