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;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-ffloat-store")
#pragma GCC optimize("-fno-defer-pop")
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
long long dx[4] = {-1, 0, 0, +1};
long long dy[4] = {0, -1, +1, 0};
const long long LINF = 1e18;
const long long INF = 1e9;
const long long MOD = 1e9 + 7;
long long strtoll(string s) {
  stringstream str(s);
  long long number = 0;
  str >> number;
  return number;
}
string lltostr(long long n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
long long llsqrt(long long x) {
  if (x == 0 || x == 1) return x;
  long long start = 1, end = x / 2, ans;
  while (start <= end) {
    long long mid = (start + end) / 2;
    if (mid * mid == x)
      return mid;
    else if (mid <= x / mid)
      ans = mid, start = mid + 1;
    else
      end = mid - 1;
  }
  return ans;
}
vector<long long> spf;
void sieve(long long size) {
  spf.resize(size + 1);
  for (long long i = 1; i <= size; ++i) spf[i] = i;
  for (long long i = 4; i <= size; i += 2) spf[i] = 2;
  for (long long i = 3; i * i <= size; ++i) {
    if (spf[i] == i) {
      for (long long j = i * i; j <= size; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
long long power(long long a, long long b, long long m) {
  if (b == 0) return 1;
  if (b == 1) return a % m;
  long long t = power(a, b / 2, m);
  t = (t * t) % m;
  if (b & 1) t = (t * a) % m;
  return t;
}
long long modInverse(long long a, long long m) { return power(a, m - 2, m); }
void fastscan(long long &number) {
  bool negative = false;
  register long long c;
  number = 0;
  c = getchar();
  if (c == '-') {
    negative = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) number = number * 10 + c - 48;
  if (negative) number *= -1;
}
long long solve() {
  long long n, m, k;
  cin >> n >> m >> k;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) cin >> v[i];
  ;
  multiset<long long> s(v.begin(), v.end());
  for (long long i = 0; i < n - k * m; i++) s.erase(s.begin());
  long long sum = 0;
  for (auto it : s) sum += it;
  cout << sum << endl;
  long long x = 1;
  for (long long i = 0; i < n; i++) {
    if (s.find(v[i]) != s.end()) {
      s.erase(s.find(v[i]));
      if (s.size() % m == 0) {
        cout << i + 1 << " ";
        x++;
        if (x == k) exit(0);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  t = 1;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}