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.math.BigInteger;
import java.util.*;
import java.io.*;
public class A{
    public static void main(String arg[]) {
        FastScanner sc = new FastScanner(System.in);
       int n = sc.nextInt();
       int m = sc.nextInt();
       int k = sc.nextInt();
       long sum =0;
       int o =0;
       long x[] = new long [n];
       int part [] = new int [k];
       int oo =0;
       for(int i =0;i<n;++i){
           x[i] = sc.nextInt();
       }
       long xx[] = x.clone();
       Arrays.sort(x);
       long km = k*m;
       long c = n-km;
       int mm = m;
       int cc = (int)c;
       int cnt = 1;
       long kkk = x[cc];

       for(int i =1;i<km;++i){

           if(x[cc]==x[cc+i]){
               cnt++;
           }
           else {break;}
       }
        
       for(int i =0;i<n;++i)
       {
           if(xx[i]>=x[cc]){

               if(xx[i]==kkk&&cnt==0){
                   continue;
               }
               if(xx[i]==kkk){
                   --cnt;
               }

               sum+=xx[i];
               o++;
           }
           if(o==m){
               part[oo++] = i+1;
               m+=mm;
           }
       }
        System.out.println(sum);
       for(int i =0;i<k-1;++i){
           System.out.print(part[i]+" ");
       }
    }


















            public static long f(long u ){
      return  u == 1?  1 :  u *f(u-1);
            }
    public int bi(long []x,int str,int end,long looking){
        int o = (str+end)/2;
        if(x[o]>looking){
            for(int i =str;i<o;++i){
                if(x[i]>=looking){
                    return i;
                }
            }
        }
        else if (x[o]<looking){
            for(int i =o;i<end;++i){
                if(x[i]>=looking){
                    return i;
                }
            }
        }
        return o;
    }




    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream stream) {
            try {
                br = new BufferedReader(new InputStreamReader(stream));
            } catch (Exception e) {
                e.printStackTrace();
            }
        }

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

        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.valueOf(next());
        }
        double nextd() {
            return Double.valueOf(next());
        }
    }
}