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 MAX_N 100005

int N,M;
vector<int> G[MAX_N];
int w[MAX_N];

bool visited[MAX_N];

int t[MAX_N];
int cc=0;

void dfs(int pos){
  visited[pos]=true;
  t[cc]=w[pos];
  cc++;
  for(int to:G[pos]){
    if(visited[to])continue;
    dfs(to);
  }
}

int main(){
  scanf("%d %d",&N,&M);
  for(int i=0;i<N;i++){
    scanf("%d",&w[i]);
  }
  for(int i=0;i<M;i++){
    int a,b;
    scanf("%d %d",&a,&b);
    G[a].push_back(b);
    G[b].push_back(a);
  }

  int X=0;
  long long ans=0;
  vector<int> v;
  for(int i=0;i<N;i++){
    if(visited[i])continue;
    X++;
    cc=0;
    dfs(i);
    sort(t,t+cc);
    ans+=t[0];
    for(int j=1;j<cc;j++)v.push_back(t[j]);
  }
  
  if(X==1){
    cout<<0<<endl;
    return 0;
  }
  
  int Y = (X-1)*2;
  
  if(Y-X > (int)v.size()){
    cout<<"Impossible"<<endl;
    return 0;
  }

  sort(v.begin(),v.end());
  for(int i=0;i<Y-X;i++){
    ans+=v[i];
  }
  cout<<ans<<endl;
  return 0;  
}
