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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.HashMap;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author MJ
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();

        solver.solve(1, in, out);
        out.close();
    }

    static class TaskB {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            PriorityQueue<Integer> pq = new PriorityQueue<>();
            int n = in.nextInt(), m = in.nextInt(), k = in.nextInt();
            ArrayList<Integer> list = new ArrayList<>();

            int topk = m * k;
            for(int i = 0; i < topk; i++) {
                int nums = in.nextInt();
                list.add(nums);
                pq.add(nums);
            }
            for(int i = topk; i < n; i++) {
                int num = in.nextInt();
                list.add(num);
                if(pq.peek() < num) {
                    pq.poll();
                    pq.add(num);
                }
            }
            HashMap<Integer, Integer> map = new HashMap<>();
            Long ans = 0L;
            for(Object i : pq.toArray()) {
                ans += new Long((Integer) i);

                map.merge((Integer) i, 1, Integer::sum);

            }

            out.println(ans);
            out.flush();
            int prn = 0;
            int kk = m;
            for(int i = 0; i < list.size(); i++) {

                if(map.get(list.get(i)) != null && map.get(list.get(i)) != 0) {
                    map.put(list.get(i), map.get(list.get(i)) - 1);
                    kk--;
                }
                if(kk == 0) {
                    out.print((i + 1) +" ");
                    kk = m;
                    prn++;
                    if(prn == k-1) break;
                }
            }

        }

    }
}