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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.math.*;

public class Main {
    public static void main(String args[]) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer stk  = null;
        stk = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(stk.nextToken());
        int m = Integer.parseInt(stk.nextToken());
        int k = Integer.parseInt(stk.nextToken());
        stk = new StringTokenizer(br.readLine());

        int[] arr = new int[n];
        Idx[] arr2 = new Idx[n];
        for(int i=0;i<n;i++){
            arr[i] = Integer.parseInt(stk.nextToken());
            arr2[i] = new Idx(arr[i],i);
        }
        Arrays.sort(arr2);
        boolean[] check  =new boolean[n];
        for(int i=n-1,cnt=0;cnt<m*k;cnt++,i--){
            check[arr2[i].idx] = true;
        }
        long sum = 0 ;
        int cnt =0 ;
        int kk = 0;
        int i=0;
        StringBuilder sb = new StringBuilder();
        for(i=0;i<n && kk < k;i++){
            if(check[i]){
                cnt++;
                sum += arr[i];
                if(cnt == m){
                    kk++;
                    if(kk!=k)
                    sb.append(i+1).append(' ');
                    cnt = 0;

                }
            }
        }
        for(;kk<k-1;kk++,i++){
            sb.append(i+1).append(' ');
        }
            System.out.println(sum);
        System.out.println(sb);
    }
    public static class Idx implements Comparable<Idx>{
        int value,idx;
        public Idx(int value,int idx){
            this.value = value;
            this.idx = idx;
        }
        @Override
        public int compareTo(Idx o) {
            if(value == o.value) return idx -o.idx;
            return value - o.value;
        }
    }
}