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 com.sun.org.apache.xpath.internal.operations.Plus;

import java.io.*;
import java.util.*;


public class Main {
    public static void main(String[] args)  {
        __SolutionBase begin = new __SolutionBase();
        begin.Solve();
    }
}

class __SolutionBase {

    public void Solve() {
        init();
        Solution();
        out.close();
    }

    private void Solution() {

        int n = readInt();
        int m = readInt();
        int k = readInt();

        long[] nums = new long[n];

        for (int i = 0;i < n;i++) nums[i] = readInt();

        long[] secondArr = nums.clone();

        Arrays.sort(nums);

        Map<Long,Integer> biggerNums = new HashMap<>();
        long answer = 0;
        for (int i = 0;i < m*k;i++) {
            if (biggerNums.containsKey(nums[n - i - 1]))
                biggerNums.put(nums[n - i - 1], biggerNums.get(nums[n - i - 1]) + 1);
            else biggerNums.put(nums[n - i - 1], 1);
            answer += nums[n - i - 1];
        }
        int countMax = 0;
        int count = 0;
        StringBuilder secondAsnwer = new StringBuilder("");
        for (int i = 0;i < n && count != k - 1 ;i++){
            if (biggerNums.containsKey(secondArr[i]) && biggerNums.get(secondArr[i]) != 0){
                countMax++;
                biggerNums.put(secondArr[i], biggerNums.get(secondArr[i]) - 1);
            }

            if (countMax == m) {
                secondAsnwer.append(i + 1 + " ");
                count++;
                countMax = 0;
            }

        }

        out.println(answer);
        out.println(secondAsnwer.toString());

    }

    /**
     * readLong, readDouble and other are similar
     *
     * @return int type from input
     */
    private int readInt() {
        return Integer.parseInt(readString());
    }

    private long readLong() {
        return Long.parseLong(readString());
    }
    /**
     * Tokens are separated by space or endExclusive of line
     *
     * @return non-empty string token from input (or null if there is no any token)
     */
    private String readString() {
        while (!tok.hasMoreTokens()) {
            String nextLine = readLine();
            if (null == nextLine) return null;
            tok = new StringTokenizer(nextLine);
        }

        return tok.nextToken();
    }

    /**
     * @return whole line from input
     */
    private String readLine() {
        try {
            return in.readLine();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }
    private BufferedReader in;
    private PrintWriter out;
    private StringTokenizer tok = new StringTokenizer(" ");

    /**
     * For local testing I use pair 'input.txt'/'output.txt' in the project root
     */
    private void init() {
        try {
            in = new BufferedReader(new FileReader("input.txt"));
            out = new PrintWriter("output.txt");
        } catch (FileNotFoundException e) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        }
    }
}