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 
import math
from collections import Counter
from operator import itemgetter
import queue

def IO():
	sys.stdin=open("pyinput.txt", 'r')
	sys.stdout=open("pyoutput.txt", 'w')

def GCD(a, b):
	if(b==0): return a
	else: return GCD(b, a%b)

def LCM(a, b): return a*(b//GCD(a, b))

def scan(TYPE_1, TYPE_2=0):
	if(TYPE_1==int):
		return map(int, sys.stdin.readline().strip().split())
	elif(TYPE_1==float):
		return map(float, sys.stdin.readline().strip().split())
	elif(TYPE_1==list and TYPE_2==float):
		return list(map(float, sys.stdin.readline().strip().split()))
	elif(TYPE_1==list and TYPE_2==int):
		return list(map(int, sys.stdin.readline().strip().split()))
	elif(TYPE_1==str):
		return sys.stdin.readline().strip()
	else: print("ERROR!!!!") 



def main():
	
	n, m, k=scan(int)
	a=scan(list, int)
	x=sorted(a, reverse=True)
	x=x[:(k*m)]
	
	mp={}
	for i in x:
		if(mp.get(i, 's')=='s'):
			mp[i]=1
		else:
			mp[i]+=1

	print(sum(x))
	ans=[]
	cnt=0
	for i in range(n):
		if(mp.get(a[i], 0)==0):
			continue
		cnt+=1
		mp[a[i]]-=1
		if(cnt==m):
			ans.append(i+1)
			cnt=0

	print(*ans[:k-1])



# IO()
main()