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.*;
import java.math.*;
public class loser
{
    static class InputReader {
        public BufferedReader br;
        public StringTokenizer token;
        public InputReader(InputStream stream)
        {
            br=new BufferedReader(new InputStreamReader(stream),32768);
            token=null;
        }

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

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

        public long nextLong()
        {
            return Long.parseLong(next());
        }
    }
    static class card{
        long a;
        int i;
        public card(long a,int i)
        {
            this.a=a;
            this.i=i;
        }
    }
    static class sort implements Comparator<pair>
    {
        public int compare(pair o1,pair o2)
        {
            if(o1.a!=o2.a)
                return (int)(o1.a-o2.a)*-1;
            else
                return (int)(o1.b-o2.b)*-1;
        }
    }
    static class sort1 implements Comparator<pair>
    {
        public int compare(pair o1,pair o2)
        {
            if(o1.b!=o2.b)
                return (int)(o1.b-o2.b);
            else
            return -1;
        }
    }
    static void shuffle(int a[])
    {
        List<Integer> l=new ArrayList<>();
        for(int i=0;i<a.length;i++)
            l.add(a[i]);
        Collections.shuffle(l);
        for(int i=0;i<a.length;i++)
            a[i]=l.get(i);
    }

    /*static long gcd(long a,long b)
    {
    if(b==0)
    return a;
    else
    return gcd(b,a%b);
    }*/
    /*static boolean valid(int i,int j)
    {
    if(i<4 && i>=0 && j<4 && j>=0)
    return true;
    else
    return false;
    }*/
    static class pair{
        long a;int b;
        public pair(long a,int b)
        {
            this.a=a;
            this.b=b;
        }
    }
    public  static void main(String[] args) 
    {
        InputReader sc=new InputReader(System.in);
        int n=sc.nextInt();
        int m=sc.nextInt();
        int k=sc.nextInt();
        pair a[]=new pair[n];
        for(int i=0;i<n;i++)
        {
            long t=sc.nextLong();
            a[i]=new pair(t,i); 
        }
        Arrays.sort(a,new sort());
        long sum=0;
        for(int i=0;i<m*k;i++)
        sum+=a[i].a;
        Arrays.sort(a,0,m*k,new sort1());
        System.out.println(sum);
        for(int i=0;i<k-1;i++)
        System.out.print((a[i*m+m-1].b+1)+" ");
        System.out.println();
    }
}