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;
const int OO = 1e9;
const long long INF = 1e18;
const int irand(int lo, int hi) {
  return ((double)rand() / (RAND_MAX + 1.0)) * (hi - lo + 1) + lo;
}
const long long lrand(long long lo, long long hi) {
  return ((double)rand() / (RAND_MAX + 1.0)) * (hi - lo + 1) + lo;
}
template <typename T>
T getnum() {
  int sign = 1;
  T ret = 0;
  char c;
  do {
    c = getchar();
  } while (c == ' ' || c == '\n');
  if (c == '-')
    sign = -1;
  else
    ret = c - '0';
  while (1) {
    c = getchar();
    if (c < '0' || c > '9') break;
    ret = 10 * ret + c - '0';
  }
  return sign * ret;
}
inline void ini(int& x) { x = getnum<int>(); }
inline void scani(int& x) { scanf("%d", &x); }
const int N = 2e5 + 5;
int n, m, k;
pair<int, int> arr[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.setf(ios::fixed);
  cout.setf(ios::showpoint);
  cout.precision(10);
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i += 1) {
    cin >> arr[i].first;
    arr[i].first *= -1;
    arr[i].second = i;
  }
  sort(arr + 1, arr + n + 1);
  long long ans = 0;
  for (int i = 1; i <= m * k; i += 1) ans += arr[i].first *= -1;
  for (int i = m * k + 1; i <= n; i += 1) arr[i].first = -OO - 5;
  for (int i = 1; i <= n; i += 1) swap(arr[i].first, arr[i].second);
  sort(arr + 1, arr + n + 1);
  cout << ans << '\n';
  int fr = 1;
  int curr = 0;
  for (int i = 1; i <= n; i += 1) {
    if (arr[i].second == -OO - 5) continue;
    curr++;
    if (curr == m) {
      k--;
      if (k == 0) break;
      cout << i << " ";
      curr = 0;
    }
  }
  cout << '\n';
  return 0;
}