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:
// package Greedy;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;

public class ArrayPartitioning {
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        String line[]=br.readLine().split(" ");
        int n=Integer.parseInt(line[0]);
        int m=Integer.parseInt(line[1]);
        int k=Integer.parseInt(line[2]);
        int a[]=new int[n];
        line=br.readLine().split(" ");
        ArrayList<Integer> arr=new ArrayList<>();
        for(int i=0;i<n;i++){
            a[i]=Integer.parseInt(line[i]);
            arr.add(a[i]);
        }
        HashMap<Integer,Integer> map=new HashMap<>();
        Collections.sort(arr);
        long sum=0;
//        int count2=0;
        for(int i=arr.size()-1;i>=arr.size()-m*k;i--){
            sum+=arr.get(i);
            if(!map.containsKey(arr.get(i)))
                map.put(arr.get(i),1);
            else{
                map.put(arr.get(i),map.get(arr.get(i))+1);
            }
//            count2++;
        }
//        System.out.println(count2);
        ArrayList<Integer> indices=new ArrayList<>();
        int index=0;
        for(int i=0;i<m*k;i+=m){
            int count=0;
//            long sum2=0;
            for(int j=index;j<n;j++){
                if(map.containsKey(a[j])&&map.get(a[j])!=0){
//                    System.out.print(a[j]+" ");
//                    sum2+=a[j];
                    count++;
                    map.put(a[j],map.get(a[j])-1);
                }
                if(count==m){
                    index=j+1;
                    indices.add(index);
                    break;
                }
            }
//            System.out.println(sum2);
        }
//        System.out.println();
        System.out.println(sum);
        for(int i=0;i<k-1;i++){
            System.out.print(indices.get(i)+" ");
        }
        System.out.println();
    }
}