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.*;
import java.util.concurrent.ThreadLocalRandom;

public class Main {
    public static void main(String[] args) throws FileNotFoundException {
        ConsoleIO io = new ConsoleIO(new InputStreamReader(System.in), new PrintWriter(System.out));

//        String fileName = "C-large-practice";
//        ConsoleIO io = new ConsoleIO(new FileReader("D:\\Dropbox\\code\\practice\\jb\\src\\" + fileName + ".in"), new PrintWriter(new File("D:\\Dropbox\\code\\practice\\jb\\src\\" + fileName + ".out")));

        new Main(io).solve();
//        new Main(io).solveLocal();

        io.close();
    }

    ConsoleIO io;

    Main(ConsoleIO io) {
        this.io = io;
    }

    ConsoleIO opt;

    Main(ConsoleIO io, ConsoleIO opt) {
        this.io = io;
        this.opt = opt;
    }

    List<List<Integer>> gr = new ArrayList<>();

    long MOD = 1_000_000_007;

    public void solve() {
        int n = io.ri(), m = io.ri(), k = io.ri();
        Pair[] all = new Pair[n];
        int[] cc = new int[n];
        long res = 0;
        for(int i = 0; i< n;i++){
            int v = io.ri();
            all[i] = new Pair(v, i);
            cc[i] = v;
        }

        Arrays.sort(all, (p1, p2) -> Integer.compare(p1.a, p2.a));
        boolean[] skip = new boolean[n];

        int tot = m*k;
        for(int i = 0; i<n-tot;i++){
            skip[all[i].b] = true;
        }

        int cur = 0;
        StringBuilder sb = new StringBuilder();
        int q = 0;
        for(int i =0; i<n;i++){
            if(cur == m && q < k - 1){
                if(q != 0){
                    sb.append(' ');
                }
                sb.append(i);
                cur = 0;
                q++;
            }
            if(!skip[i]){
                res+=cc[i];
                cur++;
            }
        }

        io.writeLine(res+"");
        io.writeLine(sb.toString());
    }

}


class ConsoleIO {

    BufferedReader br;
    PrintWriter out;
    public ConsoleIO(Reader reader, PrintWriter writer){br = new BufferedReader(reader);out = writer;}
    public void flush(){this.out.flush();}
    public void close(){this.out.close();}
    public void writeLine(String s) {this.out.println(s);}
    public void writeInt(int a) {this.out.print(a);this.out.print(' ');}
    public void writeWord(String s){
        this.out.print(s);
    }
    public void writeIntArray(int[] a, int k, String separator) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < k; i++) {
            if (i > 0) sb.append(separator);
            sb.append(a[i]);
        }
        this.writeLine(sb.toString());
    }
    public void writeLongArray(long[] a, int k, String separator) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < k; i++) {
            if (i > 0) sb.append(separator);
            sb.append(a[i]);
        }
        this.writeLine(sb.toString());
    }
    public int read(char[] buf, int len){try {return br.read(buf,0,len);}catch (Exception ex){ return -1; }}
    public String readLine() {try {return br.readLine();}catch (Exception ex){ return "";}}
    public long[] readLongArray() {
        String[]n=this.readLine().trim().split("\\s+");long[]r=new long[n.length];
        for(int i=0;i<n.length;i++)r[i]=Long.parseLong(n[i]);
        return r;
    }
    public int[] readIntArray() {
        String[]n=this.readLine().trim().split("\\s+");int[]r=new int[n.length];
        for(int i=0;i<n.length;i++)r[i]=Integer.parseInt(n[i]);
        return r;
    }
    public int[] readIntArray(int n) {
        int[] res = new int[n];
        char[] all = this.readLine().toCharArray();
        int cur = 0;boolean have = false;
        int k = 0;
        boolean neg = false;
        for(int i = 0;i<all.length;i++){
            if(all[i]>='0' && all[i]<='9'){
                cur = cur*10+all[i]-'0';
                have = true;
            }else if(all[i]=='-') {
                neg = true;
            }
            else if(have){
                res[k++] = neg?-cur:cur;
                cur = 0;
                have = false;
                neg = false;
            }
        }
        if(have)res[k++] = neg?-cur:cur;
        return res;
    }
    public int ri() {
        try {
            int r = 0;
            boolean start = false;
            boolean neg = false;
            while (true) {
                int c = br.read();
                if (c >= '0' && c <= '9') {
                    r = r * 10 + c - '0';
                    start = true;
                } else if (!start && c == '-') {
                    start = true;
                    neg = true;
                } else if (start || c == -1) return neg ? -r : r;
            }
        } catch (Exception ex) {
            return -1;
        }
    }
    public long readLong() {
        try {
            long r = 0;
            boolean start = false;
            boolean neg = false;
            while (true) {
                int c = br.read();
                if (c >= '0' && c <= '9') {
                    r = r * 10 + c - '0';
                    start = true;
                } else if (!start && c == '-') {
                    start = true;
                    neg = true;
                } else if (start || c == -1) return neg ? -r : r;
            }
        } catch (Exception ex) {
            return -1;
        }
    }
    public String readWord() {
        try {
            boolean start = false;
            StringBuilder sb = new StringBuilder();
            while (true) {
                int c = br.read();
                if (c!= ' ' && c!= '\r' && c!='\n' && c!='\t') {
                    sb.append((char)c);
                    start = true;
                } else if (start || c == -1) return sb.toString();
            }
        } catch (Exception ex) {
            return "";
        }
    }
    public char readSymbol() {
        try {
            while (true) {
                int c = br.read();
                if (c != ' ' && c != '\r' && c != '\n' && c != '\t') {
                    return (char) c;
                }
            }
        } catch (Exception ex) {
            return 0;
        }
    }
//public char readChar(){try {return (char)br.read();}catch (Exception ex){ return 0; }}
}
class Pair {
    public Pair(int a, int b) {this.a = a;this.b = b;}
    public int a;
    public int b;
}
class PairLL {
    public PairLL(long a, long b) {this.a = a;this.b = b;}
    public long a;
    public long b;
}
class Triple {
    public Triple(int a, int b, int c) {this.a = a;this.b = b;this.c = c;}
    public int a;
    public int b;
    public int c;
}