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;
void debug_out() { cerr << endl; }
void clock_out() {
  cerr << "\nTime Elapsed : " << 1.0 * clock() / CLOCKS_PER_SEC << " s\n";
}
void fileio() {
  freopen("/home/dwai/Desktop/cp/input.txt", "r", stdin);
  freopen("/home/dwai/Desktop/cp/output.txt", "w", stdout);
  freopen("/home/dwai/Desktop/cp/debug.txt", "w", stderr);
}
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << H;
  debug_out(T...);
}
int IT_MAX = 1 << 20;
const long long MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1);
const double ERR = 1e-10;
int main() {
  42;
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  long long n, m, k;
  cin >> n >> m >> k;
  long long arr[n], i;
  for (i = 0; i < n; i++) cin >> arr[i];
  long long arr2[n];
  copy(arr, arr + n, arr2);
  sort(arr2, arr2 + n, greater<long long>());
  long long sum = 0;
  map<long long, long long> store;
  for (i = 0; i < m * k; i++) sum += arr2[i], store[arr2[i]]++;
  vector<long long> v;
  long long cnt = 0;
  for (i = 0; i < n; i++) {
    if (store[arr[i]]) {
      store[arr[i]]--;
      cnt++;
      if (cnt == m) {
        v.emplace_back(i + 1);
        cnt = 0;
      }
    }
  }
  cnt = 0;
  cout << sum << "\n";
  auto itr = v.begin();
  while (itr != v.end() && cnt < k - 1) {
    cnt++;
    cout << *itr << " ";
    itr++;
  }
  return 0;
}