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;
typedef long long ll;
typedef priority_queue<ll, vector<ll>, greater<ll>> Queue;
const ll MOD = 1e9 + 7;
ll a[100000];
bool vis[100000];
vector<int> G[100000];
void dfs(int v, int p, Queue &q) {
    vis[v] = true;
    q.push(a[v]);
    for (int to : G[v]) {
        if (to == p) continue;
        dfs(to, v, q);
    }
}
int main() {
    int N, M;
    cin >> N >> M;
    if (M == N - 1) {
        cout << 0 << endl;
        return 0;
    }
    if (2 * (N - M - 1) > N) {
        cout << "Impossible" << endl;
        return 0;
    }
    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);
    }
    ll ans = 0;
    Queue tot;
    for (int i = 0; i < N; i++) {
        if (vis[i]) continue;
        Queue q;
        dfs(i, -1, q);
        ans += q.top(); q.pop();
        while (!q.empty()) {
            tot.push(q.top());
            q.pop();
        }
    }
    for (int i = 0; i < N - M - 2; i++) {
        ans += tot.top();
        tot.pop();
    }
    cout << ans << endl;
}