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 long long int mod = 1e9 + 7;
const long long int MOD = 1e18 + 7;
long long int modPow(long long int a, long long int b) {
  long long int res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long int MODPow(long long int a, long long int b) {
  long long int res = 1;
  a %= MOD;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
  }
  return res;
}
long long int nCr(long long int n, long long int k) {
  long long int res = 1;
  if (k > n - k) k = n - k;
  for (long long int i = 0; i < k; ++i) {
    res = (res * (n - i)) % mod;
    res = (res * modPow(i + 1, mod - 2)) % mod;
  }
  return res;
}
void Sieve(long long int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (long long int p = 2; p * p <= n; p++)
    if (prime[p] == true)
      for (long long int i = p * 2; i <= n; i += p) prime[i] = false;
}
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  int a[n], b[n];
  for (int i = 0; i < n; i++) cin >> a[i], b[i] = a[i];
  sort(b, b + n, greater<int>());
  map<int, int> mp;
  for (int i = 0; i < m * k; i++) mp[b[i]]++;
  long long int total = 0, sum = 0;
  for (int i = 0; i < m * k; i++) total += b[i];
  cout << total << "\n";
  int c = 0;
  std::vector<int> v;
  for (int i = 0; i < n; i++) {
    if (mp[a[i]] > 0) c++, mp[a[i]]--;
    if (c == m) v.push_back(i + 1), c = 0;
  }
  for (int i = 0; i < k - 1; i++) cout << v[i] << " ";
}