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 <bits/stdc++.h>
        using namespace std;
   #define int long long
const int MAX = 510000;
const int MOD = 1000000007;
int p[100000];
vector<int> v,v2,hen[100000];bool used[100000];
void dfs(int a){
used[a]=true;
v2.push_back(p[a]);
for(auto&& i :hen[a])if(!used[i])dfs(i);
}

     signed main(){
         cout << std::fixed << std::setprecision(15);
int n,m;
cin>>n>>m;
for(int i=0;i<n;i++)cin>>p[i];
int x,y;
for(int i=0;i<m;i++){
     cin>>x>>y;
     hen[x].push_back(y);
     hen[y].push_back(x);
}
fill(used,used+n,false);int t=0,ans=0;
for(int i=0;i<n;i++){
     if(used[i])continue;
     t++;
     dfs(i);
     sort(v2.begin(),v2.end());
     ans+=v2[0];

     for(int i=1;i<v2.size();i++)v.push_back(v2[i]);
     v2.clear();
}
if(t==1){
     cout<<0<<endl;
     return 0;
}
if(v.size()<t-2){
     cout<<"Impossible"<<endl;
     return 0;
}
  sort(v.begin(),v.end());
for(int i=0;i<t-2;i++)ans+=v[i];
cout<<ans<<endl;
     }

 


 






    





      

        
