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;

const int N = 1e5 + 10;

int n, m, a[N];
vector<int> g[N];

vector<int> vals;
bool vis[N];
void dfs(int u) {
    if (vis[u]) return;
    vis[u] = 1;
    vals.push_back(a[u]);
    for (auto v : g[u]) dfs(v);
}


int main() {
    cin >> n >> m;
    for (int i = 0; i < n; i++) cin >> a[i];
    int iter = m; while (iter--) {
        int u,v; cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    if (n-m == 1) return cout << 0, 0;

    bool good = n >= 2*(n-m-1);
    if (!good) return cout << "Impossible\n", 0;

    vector<int> rest;
    int deg = 2*(n-m-1);
    long long ans = 0;
    for (int u = 0; u < n; u++) if (!vis[u]) {
        vals.clear();
        dfs(u);
        sort(vals.begin(), vals.end(), greater<int>());

        ans += vals.back();
        vals.pop_back();
        deg--;
        for (auto x : vals) rest.push_back(x);
    }

    sort(rest.begin(), rest.end(), greater<int>());
    while (deg--) {
        ans += rest.back();
        rest.pop_back();
    }
    cout << ans << endl;

    return 0;
}
