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:
# n, m, k = 9, 2, 3
# a = [5, 2, 5, 2, 4, 1, 1, 3, 2]
#
# n, m, k = 6, 1, 4
# a = [4, 1, 3, 2, 2, 3]
#
# n, m, k = 2, 1, 2
# a = [-1000000000, 1000000000]

# n, m, k = 11, 2, 3
# a = [-7, -6, 6, -6, 5, -3, 0, 0, -1, -3, -2]

# n, m, k = 17, 1, 4
# a = [40654398, -73772996, -301750550, 47084378, -989570362, 26931632,
#      -630673783, 889561285, -392590161, 977684923, -409889397,
#      -706593539, 636930801, 546359196, 317700410, 90744216, -636840634]

n, m, k = [int(x) for x in input().strip().split(' ')]
a = [int(x) for x in input().strip().split(' ')]

limit = sorted(a)[n-m*k]

index = 0
drop = 0
while index < n and drop < n-m*k:
    if a[index] < limit:
        drop += 1
        a[index] = None
    index += 1

index = 0
while index < n and drop < n-m*k:
    if a[index] == limit:
        drop += 1
        a[index] = None
    index += 1
# print(a)

index = 0
segment = []
all = 0
while index < n:
    sub = 0
    total = 0
    while sub < m and index < n:
        if a[index] is not None:
            sub += 1
            total += a[index]
        index += 1
    while index < n and a[index] is None:
        index += 1
    all += total
    segment.append(index)
# print(segment)
print(all)
print(' '.join([str(x) for x in segment[:-1]]))


if __name__ == '__main__':
    pass