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:
/**
 * @author Finn Lidbetter
 */
import java.util.*;
import java.io.*;
import java.awt.geom.*;

public class TaskB {
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringBuilder sb = new StringBuilder();
    
    String[] s = br.readLine().split(" ");
    int n = Integer.parseInt(s[0]);
    int m = Integer.parseInt(s[1]);
    int k = Integer.parseInt(s[2]);
    long[] arr = new long[n];
    s = br.readLine().split(" ");
    long[] sorted = new long[n];
    HashMap<Long, Integer> valToCount = new HashMap<>();
    for (int i=0; i<n; i++) {
      arr[i] = Integer.parseInt(s[i]);
      sorted[i] = arr[i];
    }
    Arrays.sort(sorted);
    long sum = 0;
    for (int i=n-k*m; i<n; i++) {
      sum += sorted[i];
      if (valToCount.containsKey(sorted[i])) {
        valToCount.put(sorted[i], valToCount.get(sorted[i])+1);
      } else {
        valToCount.put(sorted[i], 1);
      }
    }
    int index = 0;
    int part = 0;
    sb.append(sum+"\n");
    while (index<n && part<k-1) {
      int currSeen = 0;
      while (currSeen<m) {
        if (valToCount.containsKey(arr[index]) && valToCount.get(arr[index])>0) {
          currSeen++;
          valToCount.put(arr[index], valToCount.get(arr[index])-1);
        }
        index++;
      }
      part++;
      sb.append((index)+" ");
    }
    sb.delete(sb.length()-1, sb.length());
    sb.append("\n");
    System.out.print(sb);
  }
}