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 Main{
static int mod = (int)(Math.pow(10, 9) + 7);
   public static void main(String[] args) {
      MyScanner sc = new MyScanner();
      out = new PrintWriter(new BufferedOutputStream(System.out));
      int n = sc.nextInt();
      int m = sc.nextInt();
      int k = sc.nextInt();
      Pair[] a = new Pair[n];
      for (int i =0 ;i < n; i++) {
    	  a[i] = new Pair(sc.nextInt(), i);
      }
      HashSet<Integer> maxI = new HashSet<Integer>();
      Arrays.sort(a);
      long maxSum = 0;
      for (int i = 0; i < m*k; i++) {
    	  maxI.add(a[i].index);
    	  maxSum += a[i].val;
      }
      
      List<Integer> breaks = new ArrayList<Integer>();
      int count = 0;
      for (int i = 0; i < n; i++) {
    	  if (maxI.contains(i)) {
    		  count++;
    	  }
    	  if (count == m) {
    		  count =0;
    		  breaks.add(i+1);
    		 // out.println(i + " asd");
    	  }
      }
      if (breaks.size() > 0) {
    	  breaks.remove(breaks.size()-1);
      }
      out.println(maxSum);
      for (int i: breaks) out.print(i + " ");
      out.println();
      out.close();
   }
   static class Pair implements Comparable<Pair>{
	   int val;
	   int index;
	   public Pair(int x, int y) {
		   val =x;
		   index=y;
	   }
	   public int compareTo(Pair o) {
		   return o.val - val;
	   }
   }
   
   static long pow(long a, long N) {
	   if (N == 0) return 1;
	   else if (N == 1) return a;
	   else {
		   long R = pow(a,N/2);
		   if (N % 2 == 0) {
			   return R*R;
		   }
		   else {
			   return R*R*a;
		   }
	   }
   }
   
   static long powMod(long a, long N) {
	   if (N == 0) return 1;
	   else if (N == 1) return a % mod;
	   else {
		   long R = powMod(a,N/2) % mod;
		   R *= R % mod;
		   if (N % 2 == 1) {
			   R *= a % mod;
		   }
			   return R % mod;
	   }
   }
static void mergeSort(int[] A){ // low to hi sort, single array only
      int n = A.length;
      if (n < 2) return;
      int[] l = new int[n/2];
      int[] r = new int[n - n/2];
 
      for (int i = 0; i < n/2; i++){
        l[i] = A[i];
      }
 
      for (int j = n/2; j < n; j++){
        r[j-n/2] = A[j];
      }
 
      mergeSort(l);
      mergeSort(r);
      merge(l, r, A);
    }
 
    static void merge(int[] l, int[] r, int[] a){
      int i = 0, j = 0, k = 0;
      while (i < l.length && j < r.length && k < a.length){
        if (l[i] < r[j]){
          a[k] = l[i];
          i++;
        }
        else{
          a[k] = r[j];
          j++;
        }
        k++;
      }
      while (i < l.length){
        a[k] = l[i];
        i++;
        k++;
      }
 
      while (j < r.length){
        a[k] = r[j];
        j++;
        k++;
      }
    }

   

   //-----------PrintWriter for faster output---------------------------------
   public static PrintWriter out;
      
   //-----------MyScanner class for faster input----------
   public static class MyScanner {
      BufferedReader br;
      StringTokenizer st;
 
      public MyScanner() {
         br = new BufferedReader(new InputStreamReader(System.in));
      }
 
      String next() {
          while (st == null || !st.hasMoreElements()) {
              try {
                  st = new StringTokenizer(br.readLine());
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
          return st.nextToken();
      }
 
      int nextInt() {
          return Integer.parseInt(next());
      }
 
      long nextLong() {
          return Long.parseLong(next());
      }
 
      double nextDouble() {
          return Double.parseDouble(next());
      }
 
      String nextLine(){
          String str = "";
	  try {
	     str = br.readLine();
	  } catch (IOException e) {
	     e.printStackTrace();
	  }
	  return str;
      }

   }
   //--------------------------------------------------------
}