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:
def main():
    # accept n,m,k
    [n,m,k]=list(map(int,input().split(" ")))
    ar=list(map(int,input().split(" ")))
    # print(ar.sort(reverse=True))
    orderedAr=ar[:]
    orderedAr.sort(reverse=True)
    items=m*k
    # print(ar)
    # print(orderedAr)
    new=orderedAr[:items]
    # make a map 
    f=dict()
    for i in new:
        if i not in f:
            f[i]=1
        else:
            f[i]+=1
    # print(f)
    # print(len(new))

    # m=dict()
    # for i in new:
        # m[]
    # Now find the partitions plus the max sum 
    # iterate thorugh the array
    j=0 
    maxSum=sum(new)
    noOfpartitions=0
    partitions=[]
    # flag=0
    # print(len(ar))
    for i in range(len(ar)):
        # print("Element being examined = ",ar[i]," ",i)
        # print("j=",j)
        if noOfpartitions==k-1:
            # print("RUN")
            # maxSum+=sum(new)
            # flag=1
            break
        elif ar[i] in f and j<m:
            if f[ar[i]]>0:
            # print(ar[i])
            # maxSum+=ar[i]
                f[ar[i]]-=1
            # some=new.pop(new.index(ar[i]))
            # removed the element
            # print(some)
                j+=1
        if j==m:
            # print("partition created")
            # print("i==",i)
            partitions.append(i+1)
            # i-=1
            # print(/"i=",i)
            noOfpartitions+=1
            j=0
        # print(new)
    print(maxSum)
    # print(partitions)
    # partitions.reverse()
    # print(partitions)
    # if flag==0:
        # partitions.append(len(ar)-1)
    # if len(partitions)!=k-1:
        # partitions.append(len(ar)-1)
    for i in partitions:
        print(i,end=" ")
main()