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 <vector>
#include <queue>
using namespace std;
typedef pair<double,double> P;
typedef long long ll;

const int mnn=1e5+1; int n,m;
priority_queue<ll, vector<ll>, greater<ll>> al, que[mnn];
ll a[mnn]; bool u[mnn]; vector<int> v[mnn];

void dfs(int from, int id) {
  //cout<<from<<" "<<id<<endl;
  que[id].push(a[from]);
  u[from]=true;
  for(int to:v[from]) {
    if (!u[to]) dfs(to, id);
  }
}

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;
    v[x].push_back(y); v[y].push_back(x);
  }
  int id = 0;
  for(int i=0;i<n;i++) {
    if (!u[i]) {dfs(i,id); id++;}
  }
  if (2 * (id-1) > n) return puts("Impossible") * 0;
  if (id == 1) return puts("0") * 0;
  ll ans=0;
  for(int i=0;i<id;i++) {
    ans += que[i].top(); que[i].pop();
    while(!que[i].empty()) {al.push(que[i].top()); que[i].pop();}
  }
  for(int i=0;i<id-2;i++) {
    ans += al.top(); al.pop();
  }
  cout<<ans<<endl;
}