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>
#define int long long
#define F first
#define S second
using namespace std;
const int maxN = 1e5 + 13, INF = 1e9 + 13;
int n, m, cmp, a[maxN], par[maxN], indx[maxN];
long long ans;
vector<int> vec[maxN], sec;
set<pair<int, int>> st;
int get_par (int x) {
    if (x == par[x])
	return x;
    return par[x] = get_par(par[x]);
}

int32_t main () {
    cin >> n >> m;
    iota(par, par + n, 0);
    for (int i = 0; i < n; i++)
	cin >> a[i];
    
    for (int i = 0; i < m; i++) {
	int x, y;
	cin >> x >> y;
	int p1 = get_par(x), p2 = get_par(y);
	if (p1 != p2) {
	    par[p1] = p2;
	}
    }
    
    if (m == n - 1)
	return cout << 0 << endl, 0;
    for (int i = 0; i < n; i++) {
	int p = get_par(i);
	vec[p].push_back(a[i]);
    }
    for (int i = 0; i < n; i++) {
	sort(vec[i].begin(), vec[i].end());
	if (vec[i].size())
	    ans += vec[i][0];
	for (int j = 1; j < vec[i].size(); j++)
	    sec.push_back(vec[i][j]);
    }

    sort(sec.begin(), sec.end());
    int s = sec.size();
    if (s < n - m - 2) 
	return cout << "Impossible" << endl, 0;
    
    for (int i = 0; i < n - m - 2; i++)
	ans += sec[i];
    cout << ans << endl;
	
	
}
	
