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 Scanline {
    public static void main(String[] args) throws Exception {
        smart_init("rec.in" , "rec.out");
        int n = nextInt();
        int m = nextInt();
        int k = nextInt();
        edge a[] = new edge[n];
        for (int i = 0; i < n; i++) {
            a[i] = new edge(nextInt(), i);
        }
        Arrays.sort(a , (o1 , o2) -> o2.x - o1.x);
        Set<Integer> ghg = new HashSet<>();

        long ans = 0;
        for (int i = 0; i < m * k; i++) {
            ans += a[i].x;
            ghg.add(a[i].id);
        }

        out.println(ans);
         int cheecker = 0;
         int id = 0;
        for (int i = 0; i < n && id < k- 1; i++) {
            if (ghg.contains(i)){
                cheecker++;
                if (cheecker == m){
                    out.print((i+1)+" ");
                    id++;
                    cheecker = 0;
                }
            }
        }
        out.close();
    }
    static BufferedReader scan;
    static PrintWriter out;
    static StringTokenizer st;
    static void init() {
        scan = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
    }
    static void init(String name1, String name2) throws Exception {
        scan = new BufferedReader(new FileReader(name1));
        out = new PrintWriter(name2);
    }
    static void smart_init(String n1, String n2) throws Exception {
        try {
            init(n1, n2);
        } catch (Exception e) {
            init();
        }
    }
    static String next() throws Exception {
        while (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(scan.readLine());
        return st.nextToken();
    }
    static int nextInt() throws Exception {
        return Integer.parseInt(next());
    }
    static long nextLong() throws Exception {
        return Long.parseLong(next());
    }
    static void dbg(int[] obj) {
        System.out.println(Arrays.toString(obj));
    }
}
class edge {
    int x, id;
    public edge(){
    }
    public edge(int x, int id){
        this.x = x;
        this.id = id;
    }
}