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;
int read() {
  int tot = 0, fh = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    tot = tot * 10 + c - '0';
    c = getchar();
  }
  return tot * fh;
}
void getstr(char *a) {
  char c = getchar();
  int len = 0;
  while (!((c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z') ||
           (c >= '0' && c <= '9')))
    c = getchar();
  while ((c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z') ||
         (c >= '0' && c <= '9')) {
    a[++len] = c;
    c = getchar();
  }
}
const int maxn = 4e5 + 100;
long long n, m, k, num, sum;
struct compass {
  int x, y;
} a[maxn];
bool cmp1(compass x, compass y) { return x.x < y.x; }
bool cmp2(compass x, compass y) { return x.y < y.y; }
int main() {
  n = read();
  m = read();
  k = read();
  for (int i = 1; i <= n; i++) {
    a[i].x = read();
    a[i].y = i;
  }
  sort(a + 1, a + n + 1, cmp1);
  for (int i = 1; i <= n - m * k; i++) a[i].x = 2000000000;
  for (int i = n - m * k + 1; i <= n; i++) sum += a[i].x;
  cout << sum << "\n";
  sort(a + 1, a + n + 1, cmp2);
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i].x != 2000000000) num++;
    if (num == m) {
      num = 0;
      cout << i << " ";
      cnt++;
      if (cnt == k - 1) break;
    }
  }
  return 0;
}