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 maxn = 1e5 * 4;
inline int g() {
  int x;
  cin >> x;
  return x;
}
inline long long gl() {
  long long x;
  cin >> x;
  return x;
}
inline string gs() {
  string s;
  cin >> s;
  return s;
}
template <typename T>
inline vector<T> gv(int len) {
  vector<T> a(len);
  for (auto &it : a) cin >> it;
  return a;
}
inline vector<int> gv(int len) {
  vector<int> a(len);
  for (auto &it : a) cin >> it;
  return a;
}
inline vector<pair<int, int>> gv_andPairCnt(int len) {
  vector<pair<int, int>> a(len);
  int i = 1;
  for (auto &it : a) {
    it.second = i;
    i++;
    cin >> it.first;
  }
  return a;
}
double power_log(double power, double val) { return log(val) / log(power); }
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (auto it : v) os << it << " ";
  os << "\n";
  return os;
}
template <typename T, typename H>
inline vector<pair<T, H>> gv_andBiPair(int len) {
  vector<pair<T, H>> a(len);
  for (auto &it : a) {
    cin >> it.first >> it.second;
  }
  return a;
}
template <typename T>
inline set<T> vector_to_set(const vector<T> &vec) {
  set<T> s;
  for (auto &it : vec) s.insert(it);
  return s;
}
const long long md = 1e9 + 7;
int main() {
  auto n = g(), m = g(), k = g();
  auto v = gv_andPairCnt(n);
  sort(v.rbegin(), v.rend());
  vector<int> pos;
  long long ans = 0;
  for (int i = 0; i < (m * k); i++) {
    ans += v[i].first;
    pos.push_back(v[i].second);
  }
  cout << ans << "\n";
  sort(pos.begin(), pos.end());
  int cnt = 0;
  for (int i = 0; i + 1 < pos.size(); i++) {
    cnt++;
    if (cnt == m) {
      cout << pos[i] << " ";
      cnt = 0;
    }
  }
}