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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool read(T& x) {
  static char buf[1 << 21], *p1 = buf, *p2 = buf;
  static char tc;
  static bool sign;
  for (tc = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin)),
             p1 == p2 ? EOF : *p1++),
      x = 0, sign = false;
       (tc < '0' || '9' < tc) && tc != EOF && tc != '-';
       tc = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin)),
             p1 == p2 ? EOF : *p1++))
    ;
  if (EOF == tc) return false;
  for (tc == '-'
       ? tc = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin)),
               p1 == p2 ? EOF : *p1++),
         sign = true : false;
       '0' <= tc && tc <= '9';
       tc = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin)),
             p1 == p2 ? EOF : *p1++))
    x = (x << 1) + (x << 3) + tc - '0';
  if (sign) x = -x;
  return true;
}
template <typename T>
void write(const T& x) {
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const int mod = 7 + 1e9;
const int N = 10 + 2e5;
long long arr[N];
pair<long long, int> brr[N];
int n, m, k;
int vis[N];
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &arr[i]);
    brr[i].first = arr[i];
    brr[i].second = i;
  }
  sort(brr + 1, brr + n + 1, greater<pair<long long, int> >());
  long long ans = 0;
  for (int i = m * k; i; --i) {
    ans += brr[i].first;
    vis[brr[i].second] = 1;
  }
  printf("%lld\n", ans);
  int tot = 0;
  for (int mm = m, i = 1; i <= n; ++i) {
    if (vis[i]) {
      if (--mm == 0) {
        if (++tot != k) printf("%d ", i);
        mm = m;
      }
    }
  }
  return 0;
}