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.Arrays;
import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.nio.charset.StandardCharsets.UTF_8;

public class Main {

    public static void main(String[] args) throws IOException {
        try (BufferedReader in = new BufferedReader(new InputStreamReader(System.in, UTF_8));
             BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out, UTF_8))) {
            StringTokenizer tokens = new StringTokenizer(in.readLine());
            int n = parseInt(tokens.nextToken());
            int m = parseInt(tokens.nextToken());
            int k = parseInt(tokens.nextToken());
            Pair[] values = new Pair[n];
            tokens = new StringTokenizer(in.readLine());
            for (int i = 0; i < n; i++) {
                values[i] = new Pair(i, parseLong(tokens.nextToken()));
            }
            PriorityQueue<Pair> queue = new PriorityQueue<>((o1, o2) -> Long.compare(o2.value, o1.value));
            queue.addAll(Arrays.asList(values));
            Pair[] selected = new Pair[m * k];
            for (int i = 0; i < m * k; i++) {
                selected[i] = queue.poll();
            }
            Arrays.sort(selected, Comparator.comparingInt(o -> o.pos));
            long sum = 0;
            for (Pair pair : selected) {
                sum += pair.value;
            }
            out.write(sum + "");
            out.newLine();
            for (int i = m; i < m * k; i += m) {
                out.write(i == m ? "" : " ");
                out.write((selected[i - 1].pos + 1) + "");
            }
        }
    }

    private static class Pair {

        final int pos;
        final long value;

        public Pair(int pos, long value) {
            this.pos = pos;
            this.value = value;
        }
    }
}