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 {

    public static void main(String[] args) throws IOException {

        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int n = sc.nextInt(), m = sc.nextInt(), k = sc.nextInt();
        int[] a = new int[n];
        PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
        for(int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
            pq.add(a[i]);
            while(pq.size() > m * k) pq.poll();
        }
        HashMap<Integer,Integer> map = new HashMap<>();
        while(!pq.isEmpty()) {
            map.put(pq.peek(), map.getOrDefault(pq.poll(), 0) + 1);
        }

        long sum = 0;
        int[] res = new int[k];
        int count_in_cur_partition = 0;
        int cur_partition_idx = 0;
        for(int i = 0; i < n; i++) {
            Integer count = map.get(a[i]);
            if (count != null && count != 0) {
                map.put(a[i], count - 1);
                count_in_cur_partition++;
                sum += a[i];
            }
            if(count_in_cur_partition == m) {
                res[cur_partition_idx++] = i + 1;
                count_in_cur_partition = 0;
            }
        }

        out.println(sum);
        for(int i = 0; i < k - 1; i++) {
            if (i > 0) out.print(" ");
            out.print(res[i]);
        }




        out.close();
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(FileReader fileReader) {
            br = new BufferedReader(fileReader);
        }

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

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

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