#include <bits/stdc++.h>
#pragma GCC optimize("-O3")
using namespace std;
const long long N = 1e6 + 5;
void pairsort(long long a[], long long b[], long long n) {
  pair<long long, long long> pairt[n];
  for (long long i = 0; i < n; i++) {
    pairt[i].first = a[i];
    pairt[i].second = b[i];
  }
  sort(pairt, pairt + n);
  for (long long i = 0; i < n; i++) {
    a[i] = pairt[i].first;
    b[i] = pairt[i].second;
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long isPrime(long long n) {
  if (n < 2) return 0;
  if (n < 4) return 1;
  if (n % 2 == 0 or n % 3 == 0) return 0;
  for (long long i = 5; i * i <= n; i += 6)
    if (n % i == 0 or n % (i + 2) == 0) return 0;
  return 1;
}
long long C(long long n, long long r) {
  if (r > n - r) r = n - r;
  long long ans = 1;
  for (long long i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
long long mod = 1e9 + 7;
long long modexpo(long long x, long long p) {
  long long res = 1;
  x = x % mod;
  while (p) {
    if (p % 2) res = res * x;
    p >>= 1;
    x = x * x % mod;
    res %= mod;
  }
  return res;
}
long long n;
string s;
int32_t main() {
  long long n, m, k;
  cin >> n >> m >> k;
  vector<long long> vec(n);
  for (long long i = 0; i < n; i++) cin >> vec[i];
  vector<long long> sm(n + 1);
  for (long long i = 1; i <= n; i++) sm[i] = sm[i - 1] + vec[i - 1];
  long long dp[n + 1][k + 1];
  for (long long i = 0; i <= n; i++)
    for (long long j = 0; j <= k; j++) {
      dp[i][j] = -1;
    }
  for (long long i = 0; i <= n; i++) {
    dp[i][0] = 0;
  }
  for (long long j = 1; j <= k; j++) {
    dp[0][j] = -1;
  }
  for (long long j = 1; j <= k; j++)
    for (long long i = 1; i <= n; i++) {
      if (j * m > i)
        continue;
      else {
        if (j * m == i) {
          dp[i][j] = sm[i] - sm[i - m] + dp[i - m][j - 1];
        } else {
          dp[i][j] = (dp[i - 1][j] > sm[i] - sm[i - m] + dp[i - m][j - 1]
                          ? dp[i - 1][j]
                          : sm[i] - sm[i - m] + dp[i - m][j - 1]);
        }
      }
    }
  cout << dp[n][k];
}

### 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>
#pragma GCC optimize("-O3")
using namespace std;
const long long N = 1e6 + 5;
void pairsort(long long a[], long long b[], long long n) {
  pair<long long, long long> pairt[n];
  for (long long i = 0; i < n; i++) {
    pairt[i].first = a[i];
    pairt[i].second = b[i];
  }
  sort(pairt, pairt + n);
  for (long long i = 0; i < n; i++) {
    a[i] = pairt[i].first;
    b[i] = pairt[i].second;
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long isPrime(long long n) {
  if (n < 2) return 0;
  if (n < 4) return 1;
  if (n % 2 == 0 or n % 3 == 0) return 0;
  for (long long i = 5; i * i <= n; i += 6)
    if (n % i == 0 or n % (i + 2) == 0) return 0;
  return 1;
}
long long C(long long n, long long r) {
  if (r > n - r) r = n - r;
  long long ans = 1;
  for (long long i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
long long mod = 1e9 + 7;
long long modexpo(long long x, long long p) {
  long long res = 1;
  x = x % mod;
  while (p) {
    if (p % 2) res = res * x;
    p >>= 1;
    x = x * x % mod;
    res %= mod;
  }
  return res;
}
long long n;
string s;
int32_t main() {
  long long n, m, k;
  cin >> n >> m >> k;
  vector<long long> vec(n);
  for (long long i = 0; i < n; i++) cin >> vec[i];
  vector<long long> sm(n + 1);
  for (long long i = 1; i <= n; i++) sm[i] = sm[i - 1] + vec[i - 1];
  long long dp[n + 1][k + 1];
  for (long long i = 0; i <= n; i++)
    for (long long j = 0; j <= k; j++) {
      dp[i][j] = -1;
    }
  for (long long i = 0; i <= n; i++) {
    dp[i][0] = 0;
  }
  for (long long j = 1; j <= k; j++) {
    dp[0][j] = -1;
  }
  for (long long j = 1; j <= k; j++)
    for (long long i = 1; i <= n; i++) {
      if (j * m > i)
        continue;
      else {
        if (j * m == i) {
          dp[i][j] = sm[i] - sm[i - m] + dp[i - m][j - 1];
        } else {
          dp[i][j] = (dp[i - 1][j] > sm[i] - sm[i - m] + dp[i - m][j - 1]
                          ? dp[i - 1][j]
                          : sm[i] - sm[i - m] + dp[i - m][j - 1]);
        }
      }
    }
  cout << dp[n][k];
}

```