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;
template <typename T>
inline bool upmax(T &x, T y) {
  return x < y ? (x = y, 1) : 0;
}
template <typename T>
inline bool upmin(T &x, T y) {
  return x > y ? (x = y, 1) : 0;
}
namespace MATH_CAL {
const int mod = 998244353;
inline int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
inline int sub(int a, int b) { return a - b < 0 ? a - b + mod : a - b; }
inline int mul(int a, int b) { return (long long)a * b % mod; }
inline void Add(int &a, int b) { (a += b) >= mod ? a -= mod : 0; }
inline void Sub(int &a, int b) { (a -= b) < 0 ? a += mod : 0; }
inline int qpow(int x, int n) {
  int r = 1;
  for (; n; n /= 2, x = mul(x, x))
    if (n & 1) r = mul(r, x);
  return r;
}
inline int mod_inv(int x) { return qpow(x, mod - 2); }
}  // namespace MATH_CAL
using namespace MATH_CAL;
const int MAX_N = 2e5 + 10;
int a[MAX_N];
inline bool cmp(int x, int y) { return a[x] > a[y]; }
int id[MAX_N], cnt[MAX_N];
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = (0), _ = (n); i < _; ++i) {
    scanf("%d", &a[i]);
    id[i] = i;
  }
  sort(id, id + n, cmp);
  long long ans = 0;
  for (int i = (0), _ = (m * k); i < _; ++i) {
    cnt[id[i]] = 1;
    ans += a[id[i]];
  }
  printf("%I64d\n", ans);
  int tot = 0, last = 0;
  for (int i = (0), _ = (n); i < _; ++i) {
    if (i > 0) cnt[i] += cnt[i - 1];
    if (cnt[i] - last == m) {
      last = cnt[i];
      ++tot;
      printf("%d%c", i + 1, " \n"[tot == k - 1]);
      if (tot == k - 1) break;
    }
  }
  return 0;
}