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:
//HashMap sortByValue(map);
//long comb=nCr(5,2);
//int gcd(s,l);
//Pair p=new Pair(x,y);

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

public class AirConditioner {

    InputStream is;
    PrintWriter out;
    String INPUT = "";

    void solve() {
        int n = ni();
        int m = ni();
        int k = ni();
        Pair[] p = new Pair[n];
        for (int i = 0; i < n; i++) {
            p[i] = new Pair(i, ni());
        }
        Arrays.sort(p, (p1, p2) -> {
            return p2.y - p1.y;
        });
        long sum=0;
        int [] idx=new int[m*k+1];
        idx[0]=-1;
        for (int i = 0; i < m*k; i++) {
            sum+=p[i].y;
            idx[i+1]=p[i].x;
        }
        Arrays.sort(idx);
        out.println(sum);
        for (int i = 1; i <= k-1; i++) {
            out.print(idx[i*m]+1+" ");
        }
    }

    long nCr(int n, int k) {
        long C[] = new long[k + 1];
        // nC0 is 1 
        C[0] = 1;
        for (int i = 1; i <= n; i++) {
            for (int j = Math.min(i, k); j > 0; j--) {
                C[j] = C[j] + C[j - 1];
            }
        }
        return C[k];
    }

    public int gcd(int s, int l) {
        if (s == 0) {
            return l;
        }

        return gcd(l % s, s);
    }

    public HashMap<Integer, Integer> sortByValue(HashMap<Integer, Integer> hm) {
        // Create a list from elements of HashMap 
        List<Map.Entry<Integer, Integer>> list
                = new LinkedList<>(hm.entrySet());

        //->change o1,o2 for reverseorder
        Collections.sort(list, (Map.Entry<Integer, Integer> o1, Map.Entry<Integer, Integer> o2) -> (o1.getValue()).compareTo(o2.getValue()));

        // put data from sorted list to hashmap  
        HashMap<Integer, Integer> temp = new LinkedHashMap<>();
        list.forEach((aa) -> {
            temp.put(aa.getKey(), aa.getValue());
        });
        return temp;
    }

    void run() throws Exception {
        is = System.in;//oj ? System.in : new ByteArrayInputStream(INPUT.getBytes());
        out = new PrintWriter(System.out);

        long s = System.currentTimeMillis();
        solve();
        out.flush();
        tr(System.currentTimeMillis() - s + "ms");
    }

    public static void main(String[] args) throws Exception {
        new AirConditioner().run();
    }

    private byte[] inbuf = new byte[1024];
    public int lenbuf = 0, ptrbuf = 0;

    private int readByte() {
        if (lenbuf == -1) {
            throw new InputMismatchException();
        }
        if (ptrbuf >= lenbuf) {
            ptrbuf = 0;
            try {
                lenbuf = is.read(inbuf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (lenbuf <= 0) {
                return -1;
            }
        }
        return inbuf[ptrbuf++];
    }

    private boolean isSpaceChar(int c) {
        return !(c >= 33 && c <= 126);
    }

    private int skip() {
        int b;
        while ((b = readByte()) != -1 && isSpaceChar(b));
        return b;
    }

    private double nd() {
        return Double.parseDouble(ns());
    }

    private char nc() {
        return (char) skip();
    }

    private String ns() {
        int b = skip();
        StringBuilder sb = new StringBuilder();
        while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    private char[] ns(int n) {
        char[] buf = new char[n];
        int b = skip(), p = 0;
        while (p < n && !(isSpaceChar(b))) {
            buf[p++] = (char) b;
            b = readByte();
        }
        return n == p ? buf : Arrays.copyOf(buf, p);
    }

    private char[][] nm(int n, int m) {
        char[][] map = new char[n][];
        for (int i = 0; i < n; i++) {
            map[i] = ns(m);
        }
        return map;
    }

    private int[] na(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = ni();
        }
        return a;
    }

    private int ni() {
        int num = 0, b;
        boolean minus = false;
        while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
        if (b == '-') {
            minus = true;
            b = readByte();
        }

        while (true) {
            if (b >= '0' && b <= '9') {
                num = num * 10 + (b - '0');
            } else {
                return minus ? -num : num;
            }
            b = readByte();
        }
    }

    private long nl() {
        long num = 0;
        int b;
        boolean minus = false;
        while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
        if (b == '-') {
            minus = true;
            b = readByte();
        }

        while (true) {
            if (b >= '0' && b <= '9') {
                num = num * 10 + (b - '0');
            } else {
                return minus ? -num : num;
            }
            b = readByte();
        }
    }

    private boolean oj = System.getProperty("ONLINE_JUDGE") != null;

    private void tr(Object... o) {
        if (!oj) {
            System.out.println(Arrays.deepToString(o));
        }
    }
}

class Pair {

    /*SORTING PAIRS BY COMPARING Y OF PAIR.
    Pair[] p=new Pair[n];
    p[i]=new Pair(i,ni());
    Arrays.sort(p,( p1, p2) -> {return p1.y-p2.y;});
     */
    int x;
    int y;

    Pair(int u, int v) {
        x = u;
        y = v;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) {
            return true;
        }
        if (!(o instanceof Pair)) {
            return false;
        }
        Pair key = (Pair) o;
        return x == key.x && y == key.y;
    }

    @Override
    public int hashCode() {
        int result = x;
        result = 31 * result + y;
        return result;
    }

}