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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.stream.Stream;

/**
 * @author madi.sagimbekov
 */
public class C1114B {

    private static BufferedReader in;
    private static BufferedWriter out;

    private static List<Integer>[] list;
    private static int[] arr;
    private static int[][] dir = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    private static boolean[] used;

    public static void main(String[] args) throws IOException {
        open();

        arr = readInts();
        int n = arr[0];
        int m = arr[1];
        int k = arr[2];

        int[] x = readInts();

        Integer[] y = new Integer[n];
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < n; i++) {
            y[i] = x[i];
        }

        Arrays.sort(y, Collections.reverseOrder());
        long result = 0;
        for (int i = 0; i < k * m; i++) {
            result += y[i];
            map.put(y[i], map.getOrDefault(y[i], 0) + 1);
        }

        List<Integer> list = new ArrayList<>();
        int cnt = 0;
        for (int i = 0; i < n; i++) {
            if (map.containsKey(x[i])) {
                int c = map.get(x[i]);
                cnt++;
                if (c == 1) {
                    map.remove(x[i]);
                } else {
                    map.put(x[i], c - 1);
                }
            }
            if (cnt == m) {
                cnt = 0;
                list.add(i + 1);
            }
        }

        out.write(result + "\n");
        for (int i = 0; i < list.size() - 1; i++) {
            out.write(list.get(i) + " ");
        }
        out.write("\n");

        close();
    }

    private static int[] readInts() throws IOException {
        return Stream.of(in.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
    }

    private static int readInt() throws IOException {
        return Integer.parseInt(in.readLine());
    }

    private static long[] readLongs() throws IOException {
        return Stream.of(in.readLine().split(" ")).mapToLong(Long::parseLong).toArray();
    }

    private static long readLong() throws IOException {
        return Long.parseLong(in.readLine());
    }

    private static double[] readDoubles() throws IOException {
        return Stream.of(in.readLine().split(" ")).mapToDouble(Double::parseDouble).toArray();
    }

    private static double readDouble() throws IOException {
        return Double.parseDouble(in.readLine());
    }

    private static String readString() throws IOException {
        return in.readLine();
    }

    private static List<Integer>[] buildAdjacencyList(int n, int m) throws IOException {
        List<Integer>[] list = new ArrayList[n + 1];
        for (int i = 0; i <= n; i++) {
            list[i] = new ArrayList<>();
        }

        for (int i = 0; i < m; i++) {
            int[] e = readInts();
            list[e[0]].add(e[1]);
            list[e[1]].add(e[0]);
        }

        return list;
    }

    private static void open() {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new BufferedWriter(new OutputStreamWriter((System.out)));
    }

    private static void close() throws IOException {
        out.flush();
        out.close();
        in.close();
    }

}