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:
#import resource
import sys
#resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
#threading.Thread(target=main).start()
#import threading
#threading.stack_size(2**26)
#sys.setrecursionlimit(10**8)
mod=(10**9)+7
#fact=[1]
#for i in range(1,1000001):
#    fact.append((fact[-1]*i)%mod)
#ifact=[0]*1000001
#ifact[1000000]=pow(fact[1000000],mod-2,mod)
#for i in range(1000000,0,-1):
#    ifact[i-1]=(i*ifact[i])%mod
from sys import stdin, stdout
from bisect import bisect_left as bl
from bisect import bisect_right as br
import itertools
import math
import heapq
from random import randint as rn
from Queue import Queue as Q
def modinv(n,p):
    return pow(n,p-2,p)
def ncr(n,r,p):
    t=((fact[n])*((ifact[r]*ifact[n-r])%p))%p
    return t
def ain():
    return map(int,sin().split())
def sin():
    return stdin.readline().strip()
def GCD(x, y):
   while(y):
       x, y = y, x % y
   return x
"""**************************************************************************"""
n,m,k=ain()
a=ain()
t=m*k
q=[]
for i in range(n):
    q.append([a[i],i])
q.sort(reverse=True)
d=[0]*n
s=0
for i in range(t):
    d[q[i][1]]+=1
    s+=q[i][0]
i=0
k1=0
r=m
s1=[]
while(k1<(k-1)):
    if(d[i]==1):
        r-=1
    if(r==0):
        s1.append(str(i+1))
        r=m
        k1+=1
    i+=1
stdout.write(str(s)+"\n")
stdout.write(" ".join(s1))