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.io.PrintStream;
import java.util.Arrays;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Wolfgang Beyer
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(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, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int m = in.nextInt();
            int k = in.nextInt();
            long[] a = in.readLongArray(n);
            long[] b = new long[n];
            System.arraycopy(a, 0, b, 0, a.length);
            Arrays.sort(a);
            long sum = 0;
            for (int i = n - m * k; i < n; i++) {
                sum += a[i];
            }
            out.println(sum);

            int count = 0;
            long cutoff = a[n - m * k];
            long skip = 0;
            int index = n - m * k - 1;
            while (index >= 0 && a[index] == cutoff) {
                skip++;
                index--;
            }

            for (int i = 0; i < n; i++) {
                if (b[i] > cutoff) count++;
                if (b[i] == cutoff) {
                    if (skip > 0) {
                        skip--;
                    } else {
                        count++;
                    }
                }
                if (count == m) {
                    count = 0;
                    if (k > 1) {
                        out.print(i + 1 + " ");
                    }
                    k--;
                }
            }

        }

    }

    static class InputReader {
        private static BufferedReader in;
        private static StringTokenizer tok;

        public InputReader(InputStream in) {
            this.in = new BufferedReader(new InputStreamReader(in));
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public long[] readLongArray(int n) {
            long[] ar = new long[n];
            for (int i = 0; i < n; i++) {
                ar[i] = nextLong();
            }
            return ar;
        }

        public String next() {
            try {
                while (tok == null || !tok.hasMoreTokens()) {
                    tok = new StringTokenizer(in.readLine());
                    //tok = new StringTokenizer(in.readLine(), ", \t\n\r\f"); //adds commas as delimeter
                }
            } catch (IOException ex) {
                System.err.println("An IOException was caught :" + ex.getMessage());
            }
            return tok.nextToken();
        }

    }
}