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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.*;

public class Codeforces {

    private static class Pair implements Comparable<Pair> {
        int index;
        long val;

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

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

    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);

        int n = in.nextInt();
        int m = in.nextInt();
        int k = in.nextInt();
        long[] arr = in.nextLongArr(n);

        List<Pair> pairs = new ArrayList<>();
        for (int i = 0; i < arr.length; i++) {
            pairs.add(new Pair(i, arr[i]));
        }
        Collections.sort(pairs);

        long res = 0;
        List<Integer> indexes = new ArrayList<>();
        for (int i = 0; i < m * k; i++) {
            int index = pairs.get(i).index;
            indexes.add(index);

            res += arr[index];
        }

        Collections.sort(indexes);
        List<Integer> resList = new ArrayList<>();
        for (int i = m - 1; i < indexes.size() - 1; i += m) {
            resList.add(indexes.get(i));
        }

        out.println(res);
        for (int i = 0; i < resList.size(); i++) {
            out.print((resList.get(i) + 1) + " ");
        }
        out.println();
        out.close();
    }

    public static int gcd(int a, int b) {
        if (b == 0) return a;
        return gcd(b, a % b);
    }

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

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
            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 String nextLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }

            return null;
        }

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

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

        public int[] nextIntArr(int n) {
            int[] arr = new int[n];
            for (int j = 0; j < arr.length; j++) {
                arr[j] = nextInt();
            }

            return arr;
        }

        public long[] nextLongArr(int n) {
            long[] arr = new long[n];
            for (int j = 0; j < arr.length; j++) {
                arr[j] = nextLong();
            }

            return arr;
        }
    }
}