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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;
import java.io.BufferedReader;
import java.io.InputStreamReader;

public class First {

    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskA solver = new TaskA();
        int t;
        //t = in.nextInt();
        t = 1;
        while (t > 0) {
            solver.call(in,out);
            t--;
        }
        out.close();

    }

    static class TaskA {
        public void call(InputReader in, PrintWriter out) {
            int n, m, k;
            long ans = 0;
            n = in.nextInt();
            m = in.nextInt();
            k = in.nextInt();

            answer[] arr = new answer[n];
            int[] array = new int[n];
            int[] ans1 = new int[k-1];

            for (int i = 0; i < n; i++) {
                arr[i] = new answer(in.nextInt(),i);
            }

            Arrays.sort(arr);

            for (int i = 0, j = 0; i < n && j< m*k; i++ ,j++) {
                array[arr[i].b] = 1;
                ans += arr[i].a;
            }

            int b = 0;

            for (int i = 0, j = 0; i < n; i++) {
                if(array[i]!=0){
                    b++;
                }

                if(b==m){
                    if(j!=k-1)
                        ans1[j] = i+1;
                    b = 0;
                    j++;
                }
            }

            out.println(ans);

            for (int i = 0; i < ans1.length; i++) {
                out.print(ans1[i]+" ");
            }
            out.println();

        }
    }

    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }

    static class answer implements Comparable<answer>{
        int a;
        int b;

        public answer(int a, int b) {
            this.a = a;
            this.b = b;
        }

        @Override
        public int compareTo(answer o) {
            return o.a - this.a;
        }
    }

    static class answer1 implements Comparable<answer1>{
        int a, b, c;

        public answer1(int a, int b, int c) {
            this.a = a;
            this.b = b;
            this.c = c;

        }

        @Override
        public int compareTo(answer1 o) {
            return this.a - o.a;
        }
    }

    static class arrayListClass {
        ArrayList<Integer> arrayList2 ;

        public arrayListClass(ArrayList<Integer> arrayList) {
            this.arrayList2 = arrayList;
        }
    }

    static long gcd(long a, long b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    static void sort(long[] a) {
        ArrayList<Long> l=new ArrayList<>();
        for (long i:a) l.add(i);
        Collections.sort(l);
        for (int i=0; i<a.length; i++) a[i]=l.get(i);
    }

    static final Random random=new Random();

    static void shuffleSort(int[] a) {
        int n=a.length;
        for (int i=0; i<n; i++) {
            int oi=random.nextInt(n), temp=a[oi];
            a[oi]=a[i]; a[i]=temp;
        }
        Arrays.sort(a);
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong(){
            return Long.parseLong(next());
        }
        public double nextDouble() {
            return Double.parseDouble(next());
        }

    }
}