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

public class b
{
   public static void main(String[] Args) throws Exception
   {
      FS sc = new FS(System.in);
      PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
      
      int n = sc.nextInt();
      int size = sc.nextInt();
      int count = sc.nextInt();
      Pair[] ps = new Pair[n];
      
      for (int i = 0; i < n; i++)
      {
         ps[i] = new Pair();
         ps[i].pos = i;
         ps[i].val = sc.nextInt();
      }
      
      Arrays.sort(ps);
      
      ArrayList<Integer> things = new ArrayList<Integer>();
      long ans = 0;
      for (int i = 0; i < size * count; i++)
      {
         things.add(ps[i].pos);
         ans += ps[i].val;
      }
      
      Collections.sort(things);
      out.println(ans);
      for (int i = 0; i < count - 1; i++)
         out.print((things.get((i + 1) * size - 1) + 1) + " ");
      
      
      out.close();
   }
   
   public static class Pair implements Comparable<Pair>
   {
      int val;
      int pos;
      
      public int compareTo(Pair o)
      {
         return o.val - val;
      }
   }
   
   public static class FS
   {
      StringTokenizer st;
      BufferedReader br;
      
      FS(InputStream in) throws Exception
      {
         br = new BufferedReader(new InputStreamReader(in));
         st = new StringTokenizer("");
      }
      
      String next() throws Exception
      {
         if (st.hasMoreTokens())
         {
            return st.nextToken();
         }
         st = new StringTokenizer(br.readLine());
         return next();
      }
      
      int nextInt() throws Exception
      {
         return Integer.parseInt(next());
      }
   }
}