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 copy
def main():
    xt = list(map(int,input().split(' ')))
    arr = list(map(int,input().split(' ')))
    if xt[0]/xt[2] == xt[1] and xt[0] % xt[2] == 0:
        i, m = 0, xt[1]
        beauty = []
        for i in range(0,len(arr),m):
            # sorted(arr[i:i+m], reverse = True)
            arr[i:i+m].sort()
            sum1 = 0
            for i in range(i, i+m, 1):
                sum1 += arr[i]
            beauty.append(sum1)
        print(sum(beauty))
        for i in range(0,len(arr), m):
            if i != 0:
                print(i, end = ' ')
    else:
        total = xt[1] * xt[2]
        left = len(arr) - total
        newArr = copy.deepcopy(arr)
        newArr.sort()
        newArr = newArr[:left]
        di = {}
        for i in newArr:
            if i not in di:
                di[i] = 0
            di[i] += 1
            
        index, skip, i = [], 1, 0
        beauty = []
        while i < len(arr):
            if len(beauty) == xt[2]:
                break;
            index.append(i)
            sum1, scount = 0, 0
            while i < len(arr):
                if arr[i] in di and di[arr[i]] != 0:
                    di[arr[i]] -= 1
                else:
                    sum1 += arr[i]
                    scount += 1
                i += 1
                if scount == xt[1]:
                    beauty.append(sum1)
                    break
                
        print(sum(beauty))
        for i in index:
            if i != 0:
                print(i, end = ' ')
        
main()