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 class Pair{
        int id;
        int val;
        Pair(int id, int val){
            this.id = id;
            this.val = val;
        }
    }

    public static void main(String[] args) {
        FastReader in = new FastReader(new BufferedInputStream(System.in));
//        Scanner in = new Scanner(new BufferedInputStream(System.in));
        PrintStream out = System.out;
        int n = in.nextInt(), m = in.nextInt(), k = in.nextInt();
        int[] arr = new int[n];
        Pair[] pairs = new Pair[n];
        for(int i = 0; i < n; i++){
            arr[i] = in.nextInt();
            pairs[i] = new Pair(i, arr[i]);
        }
        Arrays.sort(pairs, (o1, o2) -> o2.val - o1.val);
        boolean[] toUse = new boolean[n];
        long sum = 0;
        for(int i = 0; i < m * k; i++){ // m * k <= n
            toUse[pairs[i].id] = true;
            sum += pairs[i].val;
        }
        out.println(sum);
        int cnt = 0, found = 0, pos = 0;
        int[] rid = new int[k - 1];
        for(int i = 0; i < n; i++){
            cnt += toUse[i] ? 1 : 0;
            if(cnt == m){
                rid[pos++] = i + 1;
                cnt = 0;
                found++;
                if(found == k - 1) break;
            }
        }
        for(int i = 0; i < k - 1; i++)
            out.print(rid[i] + " ");
        out.println();
    }

    static class FastReader {
        public BufferedReader br;
        public StringTokenizer token;

        public FastReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream), 32768);
            token = null;
        }

        public String next() {
            while (token == null || !token.hasMoreTokens()) {
                try {
                    token = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return token.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }

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