#include <bits/stdc++.h>
using namespace std;
int n, m, k;
long long sum[5555], st[5555];
long long dp[5555][5555];
int a[5555];
long long solve(int i, int took) {
  if (i == n + 1) {
    if (took != k) return numeric_limits<int>::min();
    return 0;
  }
  long long &ret = dp[i][took];
  if (ret != -1) return ret;
  ret = solve(i + 1, took);
  if (took != k && st[i] != -1) {
    ret = max(ret, solve(i + m, took + 1) + st[i]);
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &k);
  a[0] = 0;
  for (int i = 0; i < n; i++) scanf("%d", &a[i + 1]);
  sum[0] = 0;
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i];
  for (int i = 1; i <= n; i++) {
    int ed = i + m - 1;
    if (ed > n)
      st[i] = -1;
    else
      st[i] = sum[ed] - sum[i - 1];
  }
  memset(dp, -1, sizeof dp);
  cout << solve(1, 0) << endl;
}

### Prompt

Create a solution in Cpp for 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;
int n, m, k;
long long sum[5555], st[5555];
long long dp[5555][5555];
int a[5555];
long long solve(int i, int took) {
  if (i == n + 1) {
    if (took != k) return numeric_limits<int>::min();
    return 0;
  }
  long long &ret = dp[i][took];
  if (ret != -1) return ret;
  ret = solve(i + 1, took);
  if (took != k && st[i] != -1) {
    ret = max(ret, solve(i + m, took + 1) + st[i]);
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &k);
  a[0] = 0;
  for (int i = 0; i < n; i++) scanf("%d", &a[i + 1]);
  sum[0] = 0;
  for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i];
  for (int i = 1; i <= n; i++) {
    int ed = i + m - 1;
    if (ed > n)
      st[i] = -1;
    else
      st[i] = sum[ed] - sum[i - 1];
  }
  memset(dp, -1, sizeof dp);
  cout << solve(1, 0) << endl;
}

```