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.*;
  

public class HelloWorld{
    class Pair implements Comparable<Pair>{
    int val; 
    int index;
     
    public int compareTo(Pair p)
        {return -(this.val - p.val);
            
        }
}
   
     public static void main(String []args){
   
        Scanner s = new Scanner(System.in);
        int n= s.nextInt();
        int m= s.nextInt();
        int k = s.nextInt();
        int a[] = new int [n];
        for(int i=0;i< n;i++)
            a[i] = s.nextInt();
        Pair sorted[] = new Pair[n];
        for(int i=0;i< n;i++)
        {HelloWorld st = new HelloWorld();
        HelloWorld.Pair pa = st.new Pair();
            sorted[i] = pa;
        sorted[i].val= a[i];
         sorted[i].index= i;
         
            
        }
        Arrays.sort(sorted);
                   //for(int i=0;i< n;i++)
            //System.out.print(sorted[i].val+ " ");
        int limit= m*k;
        boolean b[]= new boolean[n];
        for(int i=0;i< limit; i++)
        {b[sorted[i].index] = true;
               
        }
       // for(int i=0;i<n;i++)
        //System.out.print(b[i]+ " "); 
         
        int count=0;
        long sum= 0;
        ArrayList<Integer> lis= new ArrayList<Integer>();
        for(int i=0;i<n;i++)
        {if(b[i] == true) {count++;sum += a[i];}
            if(count==m)
                { count=0;
                    lis.add(i+1);}
        }
           System.out.println(sum);
        for(int i=0;i < lis.size()-1;i++)
           System.out.print(lis.get(i) + " ");

        
        
        
     }
}