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 as mt
import bisect
input=sys.stdin.readline
#t=int(input())
import collections 
t=1
p=10**9+7
def ncr_util():
    inv[0]=inv[1]=1
    fact[0]=fact[1]=1
    for i in range(2,300001):
        inv[i]=(inv[i%p]*(p-p//i))%p
    for i in range(1,300001):
        inv[i]=(inv[i-1]*inv[i])%p
        fact[i]=(fact[i-1]*i)%p

    
def solve():
    l1=[]
    for i in l:
        l1.append(i)
    l1.sort(reverse=True)
    ans=sum(l1[:m*k])
    j=0
    ans1=[]
    d={}
    for i in range(m*k):
        d[l1[i]]=d.get(l1[i],0)+1
    cnt=0
    suma,suma1=0,0
    for i in range(n):
        if d.get(l[i],0)!=0:
            d[l[i]]-=1
            cnt+=1
            #suma1+=l[i]
        if cnt==m and len(ans1)<k-1:
            ans1.append(i+1)
            cnt=0
    print(ans)
    return ans1
        
    
            
for _ in range(t):
    #n=int(input())
    #s=input()
    #n=int(input())
    #n,m=(map(int,input().split()))
    #n1=n
    #a=int(input())
    #b=int(input())
    #x,y,z=map(int,input().split())
    n,m,k=map(int,input().split())
    #n=int(input())
    #s=input()
    #s1=input()
    #p=input()
    l=list(map(int,input().split()))
    
    
    #l.sort(revrese=True)
    #l2=list(map(int,input().split()))
    #l=str(n)
    #l.sort(reverse=True)
    #l2.sort(reverse=True)
    #l1.sort(reverse=True)
    #print(ans)
    print(*solve())