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 PYTHON3):

Solution:
import sys
from collections import defaultdict
from math import sqrt

input = sys.stdin.readline


############ ---- Input Functions ---- ############
def inp():
    return (int(input()))


def inlt():
    return (list(map(int, input().split())))


def insr():
    s = input()
    return (list(s[:len(s) - 1]))


def invr():
    return (map(int, input().split()))


def s(d):
    dis = 1 + 4 * 2 * 3 * d
    ds = sqrt(dis)
    return (-1 + ds) / 6


def add(r, v, m):
    if len(r)<m:
        r.append(v)
        r.sort(reverse=True)

    else:
        if r[-1]<v:
            r.append(v)
            r.sort(reverse=True)
    return r[0:m]
def main():
    n,m,k = inlt()
    vals = inlt()

    d = sorted(enumerate(vals), reverse=True, key=lambda b:b[1])[0:m*k]
    print(sum(v[1] for v in d))
    d2 = sorted(d, key=lambda b:b[0])
    res = []
    i = 0
    for v in range(k):
        res.append(d2[i+m-1][0]+1)
        i = i+m
    print(" ".join([str(v) for v in res[:-1]]))









if __name__ == "__main__":
    # sys.setrecursionlimit(10 ** 6)
    # threading.stack_size(10 ** 8)
    # t = threading.Thread(target=main)
    # t.start()
    # t.join()
    main()