#include <bits/stdc++.h>
using namespace std;
const int N = 5e5;
long long int a[N], pre[N];
int n, m, k;
long long int dp[5002][5002];
long long int fun(int idx, int cur) {
  if (idx > n) {
    if (cur == k) return 0;
    return -1e15;
  }
  if (cur == k) return 0;
  long long int &check = dp[idx][cur];
  if (check != -1) return check;
  long long int ans1 = 0, ans2 = 0;
  if (idx + m - 1 <= n)
    ans1 = pre[idx + m - 1] - pre[idx - 1] + fun(idx + m, cur + 1);
  ans2 = fun(idx + 1, cur);
  return check = max(ans1, ans2);
}
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pre[i] = pre[i - 1] + a[i];
  }
  memset(dp, -1, sizeof dp);
  long long int ans = fun(1, 0);
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.

Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers:

[l1, r1], [l2, r2], ..., [lk, rk] (1 ≤ l1 ≤ r1 < l2 ≤ r2 < ... < lk ≤ rk ≤ n; ri - li + 1 = m), 

in such a way that the value of sum <image> is maximal possible. Help George to cope with the task.

Input

The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109).

Output

Print an integer in a single line — the maximum possible value of sum.

Examples

Input

5 2 1
1 2 3 4 5


Output

9


Input

7 1 3
2 10 7 18 5 33 0


Output

61

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 5e5;
long long int a[N], pre[N];
int n, m, k;
long long int dp[5002][5002];
long long int fun(int idx, int cur) {
  if (idx > n) {
    if (cur == k) return 0;
    return -1e15;
  }
  if (cur == k) return 0;
  long long int &check = dp[idx][cur];
  if (check != -1) return check;
  long long int ans1 = 0, ans2 = 0;
  if (idx + m - 1 <= n)
    ans1 = pre[idx + m - 1] - pre[idx - 1] + fun(idx + m, cur + 1);
  ans2 = fun(idx + 1, cur);
  return check = max(ans1, ans2);
}
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pre[i] = pre[i - 1] + a[i];
  }
  memset(dp, -1, sizeof dp);
  long long int ans = fun(1, 0);
  cout << ans << endl;
  return 0;
}

```