Problem: An array b is called to be a subarray of a if it forms a continuous subsequence of a, that is, if it is equal to a_l, a_{l + 1}, …, a_r for some l, r.

Suppose m is some known constant. For any array, having m or more elements, let's define it's beauty as the sum of m largest elements of that array. For example: 

  * For array x = [4, 3, 1, 5, 2] and m = 3, the 3 largest elements of x are 5, 4 and 3, so the beauty of x is 5 + 4 + 3 = 12.
  * For array x = [10, 10, 10] and m = 2, the beauty of x is 10 + 10 = 20.



You are given an array a_1, a_2, …, a_n, the value of the said constant m and an integer k. Your need to split the array a into exactly k subarrays such that:

  * Each element from a belongs to exactly one subarray.
  * Each subarray has at least m elements.
  * The sum of all beauties of k subarrays is maximum possible.

Input

The first line contains three integers n, m and k (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ m, 2 ≤ k, m ⋅ k ≤ n) — the number of elements in a, the constant m in the definition of beauty and the number of subarrays to split to.

The second line contains n integers a_1, a_2, …, a_n (-10^9 ≤ a_i ≤ 10^9).

Output

In the first line, print the maximum possible sum of the beauties of the subarrays in the optimal partition.

In the second line, print k-1 integers p_1, p_2, …, p_{k-1} (1 ≤ p_1 < p_2 < … < p_{k-1} < n) representing the partition of the array, in which:

  * All elements with indices from 1 to p_1 belong to the first subarray.
  * All elements with indices from p_1 + 1 to p_2 belong to the second subarray.
  * ….
  * All elements with indices from p_{k-1} + 1 to n belong to the last, k-th subarray.



If there are several optimal partitions, print any of them.

Examples

Input


9 2 3
5 2 5 2 4 1 1 3 2


Output


21
3 5 

Input


6 1 4
4 1 3 2 2 3


Output


12
1 3 5 

Input


2 1 2
-1000000000 1000000000


Output


0
1 

Note

In the first example, one of the optimal partitions is [5, 2, 5], [2, 4], [1, 1, 3, 2].

  * The beauty of the subarray [5, 2, 5] is 5 + 5 = 10. 
  * The beauty of the subarray [2, 4] is 2 + 4 = 6. 
  * The beauty of the subarray [1, 1, 3, 2] is 3 + 2 = 5. 



The sum of their beauties is 10 + 6 + 5 = 21.

In the second example, one optimal partition is [4], [1, 3], [2, 2], [3].

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;


public class Main implements Runnable {

    int maxn = (int)1e9;
    int inf = (int)1e9;
    long mod = (long)1e9+7;
    int n,m,k;

    void solve() throws Exception {
        n = in.iInt();
        m = in.iInt();
        k = in.iInt();
        int a[] = new int[n+100];

        PriorityQueue<Pair> q = new PriorityQueue<>((t1,t2)->(t2.x-t1.x));
        for (int i=1; i<=n; i++) {
            a[i] = in.iInt();
            q.add(new Pair(a[i],i));
        }

        int copyK = k*m;
        ArrayList<Integer> list = new ArrayList<>();
        long sum = 0L;
        while (copyK-->0) {
            Pair top = q.poll();
            list.add(top.y);
            sum+=top.x;
        }

        out.println(sum);
        Collections.sort(list);
        for (int i=m-1; i<list.size()-m; i+=m) {
            out.print(list.get(i) + " ");
        }


    }

    class Pair {
        int x,y;

        public Pair(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }

    String fileInName = "";

    static Throwable throwable;
    public static void main (String [] args) throws Throwable {
        Thread thread = new Thread(null, new Main(), "", (1 << 26));
        thread.start();
        thread.join();
        thread.run();
        if (throwable != null)
            throw throwable;
    }

    FastReader in;
    PrintWriter out;

    public void run() {

        try {
            if (!fileInName.isEmpty()) {
                in = new FastReader(new BufferedReader(new FileReader(fileInName+".in")));
                out = new PrintWriter(new BufferedWriter(new FileWriter(fileInName+".out")));
            } else {
                in = new FastReader(new BufferedReader(new InputStreamReader(System.in)));
                out = new PrintWriter(System.out);
            }

            solve();
        } catch(Exception e) {
            throwable = e;
        } finally {
            out.close();
        }

    }

    class FastReader {
        BufferedReader bf;
        StringTokenizer tk = null;

        public FastReader(BufferedReader bf) {
            this.bf = bf;
        }

        public Integer iInt() throws Exception {
            return Integer.parseInt(sString());
        }

        public Long lLong() throws Exception {
            return Long.parseLong(sString());
        }

        public Double dDouble() throws Exception {
            return Double.parseDouble(sString());
        }

        public String sString () throws Exception {
            if (tk==null || !tk.hasMoreTokens()) {
                tk = new StringTokenizer(bf.readLine());
            }
            if (!tk.hasMoreTokens()) return sString();
            else
                return tk.nextToken();
        }
    }
}