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 Answer {

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int k = sc.nextInt();
        ArrayList<Integer> arr = new ArrayList<>();
        int [] nums = new int[n];
        for (int i = 0; i < n; i++) {
            arr.add(sc.nextInt());
            nums[i] = arr.get(arr.size() - 1);
        }
        Collections.sort(arr, Collections.reverseOrder());
        long ans = 0;
        HashMap<Integer, Integer> hm = new HashMap<>();
        for (int i = 0; i < m * k; i++) {
            ans += arr.get(i);
            hm.put(arr.get(i), hm.getOrDefault(arr.get(i), 0) + 1);
        }
        System.out.println(ans);
        int ms = 0;
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (hm.containsKey(nums[i])) {
                ms++;
                hm.put(nums[i], hm.get(nums[i]) - 1);
                if (hm.get(nums[i]) == 0) {
                    hm.remove(nums[i]);
                }
            }
            if (ms == m) {
                System.out.print((i + 1) + " ");
                ms = 0;
                count++;
            }
            if (count == k - 1) break;
        }
    }

    static class Pair implements  Comparable<Pair>{
        int start;
        int end;
        int index;

        Pair(int start, int end, int index) {
            this.start = start;
            this.end = end;
            this.index = index + 1;
        }

        @Override
        public int compareTo(Pair o) {
            if(this.start == o.start) {
                return o.end - this.end;
            }
            return this.start - o.start;
        }
    }

    static class Cach implements Comparable<Cach> {
        int end;
        int location;
        Cach(int end, int location) {
            this.end = end;
            this.location = location + 1;
        }
        @Override
        public int compareTo(Cach o) {
            return this.end - o.end;
        }
    }


    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream System) throws FileNotFoundException {
            br = new BufferedReader(new InputStreamReader(System));
//            String FILENAME = "level4.txt";
//            br = new BufferedReader(new FileReader(FILENAME));
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        public char nextChar() throws IOException {
            return next().charAt(0);
        }

        public Long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public boolean ready() throws IOException {
            return br.ready();
        }

        public void waitForInput() {
            for (long i = 0; i < 3e9; i++)
                ;
        }
    }
}