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.*;
import java.util.*;
import java.lang.*;
import java.awt.*;
import java.awt.geom.*;
import java.math.*;
import java.text.*;
import java.math.BigInteger.*;
import java.util.Arrays; 

public class   fasdfasd
{
  BufferedReader in;
  StringTokenizer as;
  long nums[][],nums2[];
  boolean nums1[];
  ArrayList < Integer > [] ar;
    Map<Integer,Integer > map = new HashMap<Integer, Integer>();
  public static void main (String[] args)
  {
    new fasdfasd  ();
  }
  
  public int GCD(int a, int b) {
    if (b==0) return a;
    return GCD(b,a%b);
  }
  
  public fasdfasd  ()
  {
    try
    {
      
      in = new BufferedReader (new InputStreamReader (System.in));
      int a = nextInt();  
      int b = nextInt();
      int c  = nextInt();
      nums = new long [a][2];
      nums1 = new boolean[a];
      long total = 0;
      for(int x = 0;x<a;x++)
      {
        nums[x][0] = nextInt();
        total += nums[x][0];
        nums[x][1] = x;
      }
      Arrays.sort(nums, (aa,bb) -> Long.compare(aa[0],bb[0]));
      for(int x = 0;x<a - b * c;x++)
      {
        total -= nums[x][0];
        nums1[(int)nums[x][1]] = true;
      }
      System.out.println(total);
      int count = 0;
      int count1 = 1;
      for(int x = 0;x<a;x++)
      {
        if(count == b)
        {
          System.out.print((x)+ " " );
          count1++;
          if(count1 == c)
            break;
          count = 0;
        }
        if(nums1[x] == false)
          count++;
      }
    }
    
    catch(IOException e)
    {
    }
  }
  
  
  
  
  
  
  
  String next () throws IOException
  {
    while (as == null || !as.hasMoreTokens ())
    {
      as = new StringTokenizer (in.readLine ().trim ());
    }
    
    
    return as.nextToken ();
  }
  
  
  
  long nextLong () throws IOException
  {
    return Long.parseLong (next ());
  }
  
  
  int nextInt () throws IOException
  {
    return Integer.parseInt (next ());
  }
  
  
  double nextDouble () throws IOException
  {
    return Double.parseDouble (next ());
  }
  
  
  String nextLine () throws IOException
  {
    return in.readLine ().trim ();
  }
}