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.StringTokenizer;

public class B {
    public static void main(String[] args) throws IOException {
        try (Input input = new StandardInput(); PrintWriter writer = new PrintWriter(System.out)) {
            int n = input.nextInt(), m = input.nextInt(), k = input.nextInt();
            class NumberAndPosition {
                final int number, position;

                NumberAndPosition(int number, int position) {
                    this.number = number;
                    this.position = position;
                }
            }
            NumberAndPosition[] a = new NumberAndPosition[n];
            for (int i = 0; i < n; i++) {
                a[i] = new NumberAndPosition(input.nextInt(), i);
            }
            Arrays.sort(a, Comparator.comparingInt(x -> -x.number));
            Arrays.sort(a, 0, m * k, Comparator.comparingInt(x -> x.position));
            long answer = 0;
            for (int i = 0; i < k * m; i++) {
                answer += a[i].number;
            }
            writer.println(answer);
            for (int i = 1; i < k; i++) {
                writer.printf("%d ", a[i * m - 1].position + 1);
            }
            writer.println();
        }
    }

    interface Input extends Closeable {
        String next() throws IOException;

        default int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        default long nextLong() throws IOException {
            return Long.parseLong(next());
        }
    }

    private static class StandardInput implements Input {
        private final BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

        private StringTokenizer stringTokenizer;

        @Override
        public void close() throws IOException {
            reader.close();
        }

        @Override
        public String next() throws IOException {
            if (stringTokenizer == null || !stringTokenizer.hasMoreTokens()) {
                stringTokenizer = new StringTokenizer(reader.readLine());
            }
            return stringTokenizer.nextToken();
        }
    }
}