#include <bits/stdc++.h>
using namespace std;
const long double PI = 3.1415926535897932384626;
long long MOD = 1000000007;
const long double EPS = 1e-9;
const int N = 3e5 + 7;
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline unsigned int setBits(unsigned int v) {
  unsigned int c;
  for (c = 0; v; ++c) v &= (v - 1);
  return c;
}
inline bool isPowerOfTwo(unsigned int v) { return (v && !(v & (v - 1))); }
bool isPalin(const string& s) {
  int i, j;
  for (i = 0, j = s.length() - 1; i <= j; ++i, --j)
    if (s[i] != s[j]) return false;
  return true;
}
long long power(long long base, unsigned long long exp, long long mod = MOD) {
  long long res = 1;
  base %= mod;
  while (exp) {
    if (exp & 1) res = (res * base) % mod;
    exp >>= 1;
    base = (base * base) % mod;
  }
  return res;
}
long long modInv(long long a, long long mod = MOD) {
  if (mod == 1) return 0;
  long long x = 1, y = 0, m = mod;
  while (a > 1) {
    long long q = a / m, t = m;
    m = a % m;
    a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += mod;
  return x;
}
inline unsigned int iabs(int x) {
  unsigned int y = x;
  return (y > 0) ? y : -y;
}
inline long long add(long long a, long long b, long long mod = MOD) {
  long long res = (a % mod + b % mod) % mod;
  return res;
}
inline long long sub(long long a, long long b, long long mod = MOD) {
  long long res = (a % mod - b % mod + mod) % mod;
  return res;
}
inline long long mult(long long a, long long b, long long mod = MOD) {
  long long res = (a % mod * b % mod) % mod;
  return res;
}
inline long long div(long long a, long long b, long long mod = MOD) {
  long long res = (a % mod * modInv(b, mod)) % mod;
  return res;
}
void openf() {}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  openf();
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> arr(n);
  for (int i = 0; i < n; ++i) cin >> arr[i];
  vector<long long> csum(n + 1);
  csum[1] = arr[0];
  for (int i = 1; i < n; ++i) csum[i + 1] = (csum[i] + arr[i]);
  vector<vector<long long>> dp(n + 1, vector<long long>(k + 1));
  for (int i = m; i <= n; ++i) {
    for (int j = 1; j <= k; ++j) {
      dp[i][j] = 0;
      if (i >= m * j) {
        dp[i][j] = max(dp[i - 1][j], dp[i - m][j - 1] + csum[i] - csum[i - m]);
      }
    }
  }
  cout << dp[n][k] << endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 long double PI = 3.1415926535897932384626;
long long MOD = 1000000007;
const long double EPS = 1e-9;
const int N = 3e5 + 7;
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline unsigned int setBits(unsigned int v) {
  unsigned int c;
  for (c = 0; v; ++c) v &= (v - 1);
  return c;
}
inline bool isPowerOfTwo(unsigned int v) { return (v && !(v & (v - 1))); }
bool isPalin(const string& s) {
  int i, j;
  for (i = 0, j = s.length() - 1; i <= j; ++i, --j)
    if (s[i] != s[j]) return false;
  return true;
}
long long power(long long base, unsigned long long exp, long long mod = MOD) {
  long long res = 1;
  base %= mod;
  while (exp) {
    if (exp & 1) res = (res * base) % mod;
    exp >>= 1;
    base = (base * base) % mod;
  }
  return res;
}
long long modInv(long long a, long long mod = MOD) {
  if (mod == 1) return 0;
  long long x = 1, y = 0, m = mod;
  while (a > 1) {
    long long q = a / m, t = m;
    m = a % m;
    a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += mod;
  return x;
}
inline unsigned int iabs(int x) {
  unsigned int y = x;
  return (y > 0) ? y : -y;
}
inline long long add(long long a, long long b, long long mod = MOD) {
  long long res = (a % mod + b % mod) % mod;
  return res;
}
inline long long sub(long long a, long long b, long long mod = MOD) {
  long long res = (a % mod - b % mod + mod) % mod;
  return res;
}
inline long long mult(long long a, long long b, long long mod = MOD) {
  long long res = (a % mod * b % mod) % mod;
  return res;
}
inline long long div(long long a, long long b, long long mod = MOD) {
  long long res = (a % mod * modInv(b, mod)) % mod;
  return res;
}
void openf() {}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  openf();
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> arr(n);
  for (int i = 0; i < n; ++i) cin >> arr[i];
  vector<long long> csum(n + 1);
  csum[1] = arr[0];
  for (int i = 1; i < n; ++i) csum[i + 1] = (csum[i] + arr[i]);
  vector<vector<long long>> dp(n + 1, vector<long long>(k + 1));
  for (int i = m; i <= n; ++i) {
    for (int j = 1; j <= k; ++j) {
      dp[i][j] = 0;
      if (i >= m * j) {
        dp[i][j] = max(dp[i - 1][j], dp[i - m][j - 1] + csum[i] - csum[i - m]);
      }
    }
  }
  cout << dp[n][k] << endl;
  return 0;
}

```