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.InputStream;
import java.io.PrintStream;
import java.util.function.Supplier;
import java.util.*;
import java.util.stream.Collectors;

public class Yap {
    static Supplier<InputStream> inputSupplier = () -> System.in;
    static Supplier<PrintStream> outSupplier = () -> System.out;

    public static void main(String[] args) {
        Scanner sc = new Scanner(inputSupplier.get());
        PrintStream out = outSupplier.get();
        int n = sc.nextInt();
        int m = sc.nextInt();
        int k = sc.nextInt();
        ArrayList<Integer> arr = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            arr.add(sc.nextInt());
        }
        
        ArrayList<Integer> sorted = new ArrayList<>();
        sorted.addAll(arr);
        Collections.sort(sorted, Collections.reverseOrder());
        
        HashMap<Integer,Integer> map = new HashMap<>();
        long max = 0;
        for (int i = 0; i < m * k; i++) {
            int curr = sorted.get(i);
            map.merge(curr, 1, (a, b) -> a + 1);
            max += curr;
        }

        ArrayList<Integer> result = new ArrayList<>();

        int currFoundNum = 0;
        int subarrNum = 0;
        for (int i = 0; i < n; i++) {
            int curr = arr.get(i);
            if (map.containsKey(curr)) {
                int num = map.get(curr);
                currFoundNum++;
                if(num>1)
                    map.put(curr, num - 1);
                else
                    map.remove(curr);

                if (currFoundNum == m) {
                    currFoundNum = 0;
                    result.add(i + 1);
                    subarrNum++;
                    if (subarrNum == (k-1)) {
                        break;
                    }
                }
            }
        }
        out.println(max);
        out.println(result.stream()
                .map(String::valueOf)
                .collect(Collectors.joining(" ")));
    }
}