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.*;

public class B {

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

    private static class Solver {

        private class Pair implements Comparable<Pair> {
            int a, b;

            public Pair(int a, int b) {
                this.a = a;
                this.b = b;
            }

            @Override
            public boolean equals(Object o) {
                if (this == o) return true;
                if (!(o instanceof Pair)) return false;
                Pair p = (Pair) o;
                return (a == p.a) && (b == p.b);
            }

            @Override
            public int hashCode() {
                return 31 * Integer.valueOf(a).hashCode() + Integer.valueOf(b).hashCode();
            }

            @Override
            public int compareTo(Pair o) {
                return o.a - a;
            }
        }

        private void solve(InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int m = in.nextInt();
            int k = in.nextInt();

            int[] a = new int[n];
            ArrayList<Pair> sorted = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                a[i] = in.nextInt();
                sorted.add(new Pair(a[i], i));
            }
            Collections.sort(sorted);

            HashSet<Pair> hs = new HashSet<>();
            for (int i = 0; i < m * k; i++) {
                hs.add(sorted.get(i));
            }

            long sum = 0;
            ArrayList<Integer> p = new ArrayList<>();
            for (int i = 0, j = 0; i < n; i++) {
                if (hs.contains(new Pair(a[i], i))) {
                    sum += a[i];
                    j++;
                    if (j % m == 0 && j != m * k) {
                        p.add(i + 1);
                    }
                }
            }

            out.println(sum);
            for (int i : p) out.print(i + " ");
            out.println();
        }
    }

    private static class InputReader {
        private BufferedReader br;
        private StringTokenizer st;

        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream));
        }

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

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

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

        private String nextLine() {
            String s = "";
            try {
                s = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return s;
        }

        private String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
    }
}