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){
        try {
            new Main().solve();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
    private void solve() throws Exception{
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int n=in.nextInt(),m=in.nextInt(),k=in.nextInt();
        Num[]a=new Num[n];
        for(int i=0;i<n;i++){
            a[i]=new Num(in.nextInt());
            a[i].id=i;
        }
        Arrays.sort(a,Comparator.comparingInt(x->x.a));
        long ans=0;
        for(int i=n-1;i>=n-k*m;i--){
            a[i].f=true;
            ans+=a[i].a;
        }
        Arrays.sort(a,Comparator.comparingInt(x->x.id));
        int cnt=0,x=0;
        out.println(ans);
        for(int i=0;i<n;i++){
            if(a[i].f)cnt++;
            if(cnt==m&&x<k-1){
                out.print(i+1+(x==k-2?"\n":" "));
                x++;
                cnt=0;
            }
        }
        out.flush();
    }

}
class Num{
    public int a,id;
    boolean f=false;
    Num(int x){
        a=x;
    }
}
class InputReader{
    StreamTokenizer tokenizer;
    public InputReader(InputStream stream){
        tokenizer=new StreamTokenizer(new BufferedReader(new InputStreamReader(stream)));
        tokenizer.ordinaryChars(33,126);
        tokenizer.wordChars(33,126);
    }
    public String next() throws IOException {
        tokenizer.nextToken();
        return tokenizer.sval;
    }
    public int nextInt() throws IOException {
        return Integer.parseInt(next());
    }
    public long nextLong() throws IOException {
        return Long.parseLong(next());
    }
    public double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }
    public boolean hasNext() throws IOException {
        int res=tokenizer.nextToken();
        tokenizer.pushBack();
        return res!=tokenizer.TT_EOF;
    }
}