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 <string>
#include <algorithm>
using namespace std;
#define rep(i,a,b) for(int i = a; i < b; i++)
#define inf 1000000000000
typedef long long ll;
typedef long double ld;
typedef vector<ll> vi;
typedef pair<ll,ll> ii;
typedef vector<ii> vii;

vector<vi> edges;
vi _,a,comp,vis,best,rest;
ll cur=0;

void dfs(ll v){
  vis[v] = 1;
  comp[v] = cur;
  rep(i,0,edges[v].size()){
    if (!vis[edges[v][i]]) { dfs(edges[v][i]); }
  }
}

int main(){
  ll N,M,x,y,res=0; cin >> N >> M; edges.assign(N,_); a.assign(N,0); vis.assign(N,0); comp.assign(N,-1);
  rep(i,0,N) cin >> a[i];
  rep(i,0,M) { cin >> x >> y; edges[x].push_back(y); edges[y].push_back(x); }

  rep(i,0,N){
    if (!vis[i]){ dfs(i); cur++; }
  }

  if (2*(cur-1) > N) { cout << "Impossible" << endl; exit(0); }

  best.assign(cur,inf);
  rep(i,0,N) {
    if (best[comp[i]] > a[i]) {
      if (best[comp[i]] != inf) rest.push_back(best[comp[i]]);
      best[comp[i]] = a[i];
    } else { rest.push_back(a[i]); }
  }
  if (cur == 1) {cout << 0 << endl; exit(0);}
  rep(i,0,best.size()) res += best[i];
  sort(rest.begin(), rest.end());
  rep(i,0,cur-2) res += rest[i];
  cout << res << endl;

}
