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.util.List;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
import java.util.stream.Collectors;

public class Main
{
    private static final FIO fio = new FIO(System.in);
    private static final StringBuilder output = new StringBuilder();

    public static void main(String args[]) throws IOException
    {
        int n, m, k;

        n = fio.readInt();
        m = fio.readInt();
        k = fio.readInt();

        PriorityQueue<Pair> heap = new PriorityQueue<Pair>((p1, p2) -> {
            if(p1.val>p2.val)
                return 1;
            else if(p1.val<p2.val)
                return -1;
            else
                return 0;
        });

        for (int x, i = 0; i < n; i++)
        {
            x = fio.readInt();
            if (heap.size() == (m * k))
            {
                if(x >= heap.peek().val)
                {
                    heap.poll();
                    heap.add(new Pair(x, i));
                }
            }
            else
            {
                heap.add(new Pair(x, i));
            }
        }

        List<Long> list = heap.stream()
                .map(p -> p.index)
                .collect(Collectors.toList());

        list.sort((x, y) -> {
            if(x>y)
                return 1;
            else if(x<y)
                return -1;
            else
                return 0;
        });

        System.out.println(heap.stream().mapToLong(x -> x.val).sum());

        for(int i=1;i<list.size();i++)
        {
            if(i%m==0)
                System.out.print(list.get(i)+" ");
        }
    }

}

class Pair
{
    long val;
    long index;

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

class FIO
{
    private BufferedReader reader;
    private StringTokenizer tokenizer;

    public FIO(InputStream stream)
    {
        reader = new BufferedReader(new InputStreamReader(stream));
    }

    public String next()
    {
        while (tokenizer == null || !tokenizer.hasMoreTokens())
        {
            try
            {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e)
            {
                throw new RuntimeException();
            }
        }
        return tokenizer.nextToken();
    }

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

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