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 com.sun.org.apache.bcel.internal.generic.BREAKPOINT;
import javafx.util.*;
import java.util.*;
import java.io.*;
import java.math.*;

public class Test10 {

    Random rnd = new Random();
    PrintWriter pw = new PrintWriter(System.out);
    int a;
    static int[] nn;

    void run() {
        int n = ni(), m = ni(), k = ni();
        long o = 0;
        nn = new int[n];
        for(int q=0; q<n; q++) nn[q] = q;
        int[] p = na(n), cl = new int[n], otm = new int[n];
        for(int q=0; q<n; q++) cl[q] = p[q];
        PyraSort.sort(cl);
        for(int q=n-1, i=0; i<m*k; q--, i++){
            otm[nn[q]] = 1;
            o+=p[nn[q]];
        }
        pw.println(o);
        int s=0;
        for(int q=0, c=0; q<n; q++){
            s+=otm[q];
            if(s==m && c+1!=k) {
                pw.print((q + 1) + " ");
                s=0; c++;
            }
        }
        pw.flush();
    }

    static class PyraSort {

        private static int heapSize;

        public static void sort(int[] a) {
            buildHeap(a);
            while (heapSize > 1) {
                swap(a, 0, heapSize - 1);
                heapSize--;
                heapify(a, 0);
            }
        }

        private static void buildHeap(int[] a) {
            heapSize = a.length;
            for (int i = a.length / 2; i >= 0; i--) {
                heapify(a, i);
            }
        }

        private static void heapify(int[] a, int i) {
            int l = 2 * i + 2;
            int r = 2 * i + 1;
            int largest = i;
            if (l < heapSize && a[i] < a[l]) {
                largest = l;
            }
            if (r < heapSize && a[largest] < a[r]) {
                largest = r;
            }
            if (i != largest) {
                swap(a, i, largest);
                heapify(a, largest);
            }
        }

        private static void swap(int[] a, int i, int j) {
            a[i] ^= a[j] ^= a[i];
            a[j] ^= a[i];
            nn[i] ^= nn[j] ^= nn[i];
            nn[j] ^= nn[i];
        }
    }

    public static void main(String[] args) {
        new Test10().run();
    }

    InputStream is = System.in;
    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 nline(){
        int b = readByte();
        StringBuilder sb = new StringBuilder();
        while (b!=10) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    private String ns() {
        int b = skip();
        StringBuilder sb = new StringBuilder();
        while (!(isSpaceChar(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();
        }
    }
}