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 javax.print.attribute.IntegerSyntax;
import java.io.*;
import java.util.*;

/**
 * @author gamezovladislav
 */

public class TaskC {
    public static void main(String[] args) {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        Solve solve = new Solve(in, out);
        solve.solve();
        out.close();
    }

    private static class Solve {

        private final InputReader in;
        private final PrintWriter out;

        public Solve(InputReader in, PrintWriter out) {
            this.in = in;
            this.out = out;
        }

        public void solve() {
            int n = in.nextInt();
            int m = in.nextInt();
            int k = in.nextInt();
            List<Pair> list = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                list.add(new Pair(in.nextLong(), i));
            }
            Collections.sort(list);
            int maxx = m*k;
            boolean[] used = new boolean[n];
            long answerSum = 0L;
            for (int i = 0; i < maxx; i++) {
                Pair pair = list.get(i);
                used[pair.index] = true;
                answerSum += pair.value;
            }


            List<Integer> length = new ArrayList<>();
            int cur = 0;
            for (int i = 0; i < n; i++) {
                if (used[i]) {
                    cur++;
                }
                if (cur == m) {
                    length.add(i + 1);
                    cur = 0;
                }
            }
//            for (boolean x : used) {
//                out.print(x + " ");
//            }
//            out.println();
            out.println(answerSum);
            length.remove(length.size() - 1);
            for (int x : length) {
                out.print(x + " ");
            }
        }

        private class Pair implements Comparable {
            private long value;
            private int index;

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


            @Override
            public int compareTo(Object o) {
                if (o instanceof Pair) {
                    if (this.value > ((Pair) o).value) return -1;
                    if (this.value < ((Pair) o).value) return 1;
                    return 0;
                }
                return -1;
            }
        }
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}