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;
long long int mod = 1e9 + 7;
const long long int MAXN = 1e6 + 1;
long long int modexp(long long int x, long long int y, long long int p) {
  if (x == 0) {
    return 0;
  }
  if (y == 0) {
    return 1;
  }
  if (y % 2 == 0) {
    long long int res = modexp(x, y / 2, p);
    return (res * res) % p;
  } else {
    long long int res = modexp(x, y - 1, p);
    return (x * res) % p;
  }
}
long long int inverse(long long int n, long long int p) {
  return modexp(n, p - 2, p);
}
long long int phi[MAXN], tot[MAXN];
void etf() {
  for (long long int i = 1; i < MAXN; i++) {
    phi[i] = i;
  }
  for (long long int i = 2; i < MAXN; i++) {
    if (phi[i] == i) {
      phi[i] = i - 1;
      for (long long int j = 2 * i; j < MAXN; j += i) {
        phi[j] = (phi[j] * (i - 1)) / i;
      }
    }
  }
  for (long long int i = 1; i < MAXN; i++) {
    for (long long int j = i; j < MAXN; j += i) {
      tot[j] += i * phi[i];
    }
  }
}
long long int fact[MAXN];
long long int ncr(long long int n, long long int r) {
  long long int ans = ((fact[n] % mod * inverse(fact[r], mod) % mod) % mod *
                       inverse(fact[n - r], mod) % mod) %
                      mod;
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m, k;
  cin >> n >> m >> k;
  long long int arr[n];
  long long int brr[n];
  for (long long int i = 0; i < n; i++) {
    cin >> arr[i];
    brr[i] = arr[i];
  }
  sort(brr, brr + n);
  map<long long int, long long int> mp;
  long long int res = 0;
  long long int check = 0;
  for (long long int i = n - 1; i >= 0; i--) {
    mp[brr[i]]++;
    res += brr[i];
    check++;
    if (check == m * k) {
      break;
    }
  }
  long long int cnt = 0;
  long long int t = 0;
  cout << res << endl;
  for (long long int i = 0; i < n; i++) {
    if (mp[arr[i]] > 0) {
      cnt++;
      mp[arr[i]]--;
    }
    if (cnt == m) {
      t++;
      cout << i + 1 << " ";
      cnt = 0;
    }
    if (t == k - 1) {
      break;
    }
  }
  return 0;
}