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.*;
public class MainClass
{
	public static void main(String args[])throws IOException
	{
	    try
	    {
    		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
    		String t_input[];
            t_input=br.readLine().split(" ");
            int n=Integer.parseInt(t_input[0]);
            int m=Integer.parseInt(t_input[1]);
            int k=Integer.parseInt(t_input[2]);
            t_input=br.readLine().split(" ");
            int a[]=new int[n];
            int b[]=new int[n];
            int c[]=new int[k-1];
            for(int i=0;i<n;i++)
            {
                a[i]=Integer.parseInt(t_input[i]);
                b[i]=a[i];
            }
            Arrays.sort(b);
            int pos=n-(m*k);
            int limit=b[pos];
            int limcount=0;
            for(int i=n-1;i>=pos;i--)
            {
                if(b[i]==limit)
                limcount++;
            }
            //System.out.println(limit+" "+limcount);
            int count=0;
            long s=0;
            int p=0;
            long max=0;
            for(int i=0;i<n;i++)
            {
                if(a[i]>limit)
                {
                    count++;
                    s+=a[i];
                }
                else if(a[i]==limit&&limcount>0)
                {
                    limcount--;
                    count++;
                    s+=a[i];
                }
                if(count>=m)
                {
                    if(p<k-1)
                    c[p++]=i+1;
                    max+=s;
                    count=0;
                    s=0;
                }
            }
            System.out.println(max);
            for(int i=0;i<p;i++)
            System.out.print(c[i]+" ");
	    }
		catch(NumberFormatException x){}
	}
}
/*
__________TEST INPUT:
int test=Integer.parseInt(br.readLine());
    	    
__________n INPUT:
int n=Integer.parseInt(br.readLine());
    		
__________ARRAY INPUT:
        String a_input[];
        a_input=br.readLine().split(" ");
        int a[]=new int[n];
        for(int i=0;i<n;i++)
        {
            a[i]=Integer.parseInt(a_input[i]);
        }

__________INPUT FOR TWO SPACE SEPARATED INTEGERS:
        String t_input[];
        t_input=br.readLine().split(" ");
        int n=Integer.parseInt(t_input[0]);
        int k=Integer.parseInt(t_input[1]);

__________STRING TO INTEGER ARRAY: ex-12345 to 1,2,3,4,5
	String st=br.readLine();
	char chars[]=st.toCharArray();
	int a[]=new int[st.length()];
	for(int i=0;i<st.length();i++)
	a[i]=chars[i]-48;

__________ARRAY WITH SPACE SEPARATED INTEGERS:
		int a[]=new int[n];
	String t_input[];
        t_input=br.readLine().split(" ");
        for(int i=0;i<n;i++)
        {
             a[i]=Integer.parseInt(t_input[i]);
        }

__________TO SORT IN DESCENDING: Arrays.sort(b,Collections.reverseOrder());
*/