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.util.*;
import java.io.*;

public class Main {

    FastReader in = new FastReader();
    PrintWriter out = new PrintWriter(System.out);
    long mod = (long)1e9+7;
    int ma = (int)1e5;

    // main code 
    void solve(){
        int n = in.ni(), m = in.ni(), k = in.ni();
        Integer[] a = new Integer[n];
        Integer[] b = new Integer[n];
        for(int i=0;i<n;i++){
            a[i] = in.ni();
            b[i] = a[i];
        }
        Arrays.sort(a,Collections.reverseOrder());
        HashMap<Integer,Integer> ts = new HashMap<>();
        long ans = 0;
        for(int i=0;i<m*k;i++){
            ts.put(a[i],ts.getOrDefault(a[i],0)+1);
            ans += a[i];
        }
        ArrayList<Integer> res = new ArrayList<>();
        int x = 0;
        for(int i=0;i<n-1 && x<k-1;i++){
            int j = 0;
            while(j<m && i<n){
                if(ts.getOrDefault(b[i],0)>0){
                    j++;
                    ts.put(b[i],ts.get(b[i])-1);
                }
                i++;
                // out.println(j);
            }
            res.add(i);
            i--;
            x++;
        }
        out.println(ans);
        for(int i : res)out.print(i+" ");
    }

    int rand(int a){
        return (int)(Math.random()*a);
    }

    int get(char a){
        if(a<='Z') return a-'A'+27;
        return a-'a'+1;
    }

    char get(int a){
        if(a<=26) return (char)(a-1+'a');
        return (char)(a-27+'A');
    }

    void run(){
        solve();
        // out.flush();
        out.close();
    }
    public static void main(String[] args) {
        new Main().run();
    }
    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next(){
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine(){
            String str = "";
            try{   
                str = br.readLine();
            }catch (IOException e){
                e.printStackTrace();
            }  
            return str;
        }
        int nextInt(){return Integer.parseInt(in.next());}
        long nextLong(){return Long.parseLong(in.next());}
        double nextDouble(){return Double.parseDouble(in.next());}
        int ni(){return Integer.parseInt(in.next());}
        long nl(){return Long.parseLong(in.next());}
        double nd(){return Double.parseDouble(in.next());}
    }
    class Edges{
        int u,v,w;
        public Edges(int u,int v,int w){
            this.u = u;
            this.v = v;
            this.w = w;
        }
    }
    void pn(Object o){out.println(o);out.flush();}
    long inv(long a, long b){return 1<a ? b - inv(b%a,a)*b/a : 1;}//a^(-1)%b
    long fe(long a,long b){
        long ans = 1L;
        while(b>0){
            if((b&1)>0)ans=ans*a%mod;
            a = a%mod*a%mod;
            b>>=1;
        }
        // pn("ans = "+ans);
        return ans;
    }
    long gcd(long a,long b){
        if(a==0)return b;
        return gcd(b%a,a);
    }
    class pair{
        int f,s;
        public pair(int f,int s){
            this.f = f;
            this.s = s;
        }
    }
}