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 P1114B {
    public static void main(String[] args) {
        InputReader in = new InputReader(System.in);
        PrintWriter pw = new PrintWriter(System.out);
//        int test = in.nextInt();
        int test = 1;
        while(test-->0) {
            int n = in.nextInt(), m = in.nextInt(), k = in.nextInt();
            long[] a = in.nextLongArray(n);
            PriorityQueue<Long> pq = new PriorityQueue<>();
            for (int i = 0; i < a.length; i++) {
                if (pq.size() < m * k) {
                    pq.add(a[i]);
                } else {
                    if (pq.peek() < a[i]) {
                        pq.poll();
                        pq.add(a[i]);
                    }
                }
            }
            HashMap<Long, Integer> cnt = new HashMap<>();
            long total = 0;
            while (!pq.isEmpty()) {
                long val = pq.poll();
                total += val;
                cnt.put(val, cnt.getOrDefault(val, 0) + 1);
            }
            pw.println(total);
            long counter = 0;
            long times = 0;
            for (int i = 0; i < a.length; i++) {
                if (cnt.containsKey(a[i])) {
                    counter++;
                    cnt.put(a[i], cnt.get(a[i]) - 1);
                    if (cnt.get(a[i]) == 0) {
                        cnt.remove(a[i]);
                    }
                }
                if (counter == m) {
                    pw.print((i + 1) + " ");
                    counter = 0;
                    times++;
                    if (times == k - 1) break;
                }
            }
        }
        pw.close();
    }

    static class InputReader {

        private final InputStream stream;
        private final byte[] buf = new byte[8192];
        private int curChar, snumChars;

        public InputReader(InputStream st) {
            this.stream = st;
        }

        public int read() {
            if (snumChars == -1)
                throw new InputMismatchException();
            if (curChar >= snumChars) {
                curChar = 0;
                try {
                    snumChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (snumChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }

        public int nextInt() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int res = 0;
            do {
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

        public long nextLong() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            long res = 0;
            do {
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

        public int[] nextIntArray(int n) {
            int [] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextInt();
            }
            return a;
        }
        public long[] nextLongArray(int n) {
            long [] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextLong();
            }
            return a;
        }

        public static int[] shuffle(int[] a, Random gen)
        { for(int i = 0, n = a.length;i < n;i++)
        { int ind = gen.nextInt(n-i)+i;
            int d = a[i];
            a[i] = a[ind];
            a[ind] = d;

        }
            return a;
        }


        public String readString() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }

        public String nextLine() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isEndOfLine(c));
            return res.toString();
        }

        public boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        private boolean isEndOfLine(int c) {
            return c == '\n' || c == '\r' || c == -1;
        }
    }
}