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 PYTHON):

Solution:
#!/usr/bin/env python
"""
This file is part of https://github.com/cheran-senthil/PyRival
Copyright 2019 Cheran Senthilkumar <hello@cheran.io>

"""
from __future__ import division, print_function

import cmath
import itertools
import math
import operator as op
# import random
import sys
from atexit import register
from bisect import bisect_left, bisect_right
from collections import Counter, defaultdict, deque
# from copy import deepcopy
# from decimal import Decimal
# from difflib import SequenceMatcher
# from functools import reduce
# from heapq import heappop, heappush
from io import BytesIO, FileIO, StringIO

if sys.version_info[0] < 3:

    class dict(dict):
        """dict() -> new empty dictionary"""

        def items(self):
            """D.items() -> a set-like object providing a view on D's items"""
            return dict.iteritems(self)

        def keys(self):
            """D.keys() -> a set-like object providing a view on D's keys"""
            return dict.iterkeys(self)

        def values(self):
            """D.values() -> an object providing a view on D's values"""
            return dict.itervalues(self)

    input = raw_input
    range = xrange

    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip

INP_FILE = 0
OUT_FILE = 1

if sys.version_info[0] < 3:
    sys.stdin = BytesIO(FileIO(INP_FILE).read())
    sys.stdout = BytesIO()
    register(lambda: FileIO(OUT_FILE, 'w').write(sys.stdout.getvalue()))
else:
    sys.stdin = StringIO(FileIO(INP_FILE).read().decode())
    sys.stdout = StringIO()
    register(lambda: FileIO(OUT_FILE, 'w').write(sys.stdout.getvalue().encode(
    )))

input = lambda: sys.stdin.readline().rstrip('\r\n')


def main():
    n, m, k = map(int, input().split())
    a = list(map(int, input().split()))

    _mk_largest = sorted(a, reverse=True)[:m * k]
    ans = sum(_mk_largest)
    mk_largest = Counter(_mk_largest)

    res = []

    i = 0
    while i < n:
        j = 0
        while (i < n) and (j < m):
            if a[i] in mk_largest:
                j += 1
                mk_largest[a[i]] -= 1

                if mk_largest[a[i]] == 0:
                    del mk_largest[a[i]]

            i += 1

        res.append(i)

    res = res[:k - 1]

    print(ans)
    print(*res)


if __name__ == '__main__':
    main()