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 <algorithm>

using namespace std;

vector<int> G[100000];
int a[100000];
bool used[100000] = {false};

int cnt;

int dfs(int n, int m){
    if(a[n] < a[m]) m = n;
    used[n] = true;
    for(int i  = 0; i < G[n].size(); i++){
        if(!used[G[n][i]]) m = dfs(G[n][i], m);
    }
    return m;
}

int main(){
    int N, M;
    long ans = 0;
    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);
        //cout << x << y << endl;
    } 
    for(int i = 0; i < N; i++){
        if(!used[i]){
            int idx = dfs(i, i);
            ans += a[idx];
            a[idx] = __INT_MAX__;
        } 
    }
    sort(a, a+N);
    for(int i = 0; i < N-M-2; i++){
        if(a[i] == __INT_MAX__){
            cout << "Impossible" << endl;
            return 0;
        }
        ans += a[i];
    }
    if(N-M == 1){
        cout << 0 << endl;
    }else{
        cout << ans << endl;
    }
}