This morning, Roman woke up and opened the browser with n opened tabs numbered from 1 to n. There are two kinds of tabs: those with the information required for the test and those with social network sites. Roman decided that there are too many tabs open so he wants to close some of them.

He decided to accomplish this by closing every k-th (2 ≤ k ≤ n - 1) tab. Only then he will decide whether he wants to study for the test or to chat on the social networks. Formally, Roman will choose one tab (let its number be b) and then close all tabs with numbers c = b + i ⋅ k that satisfy the following condition: 1 ≤ c ≤ n and i is an integer (it may be positive, negative or zero).

For example, if k = 3, n = 14 and Roman chooses b = 8, then he will close tabs with numbers 2, 5, 8, 11 and 14.

After closing the tabs Roman will calculate the amount of remaining tabs with the information for the test (let's denote it e) and the amount of remaining social network tabs (s). Help Roman to calculate the maximal absolute value of the difference of those values |e - s| so that it would be easy to decide what to do next.

Input

The first line contains two integers n and k (2 ≤ k < n ≤ 100) — the amount of tabs opened currently and the distance between the tabs closed.

The second line consists of n integers, each of them equal either to 1 or to -1. The i-th integer denotes the type of the i-th tab: if it is equal to 1, this tab contains information for the test, and if it is equal to -1, it's a social network tab.

Output

Output a single integer — the maximum absolute difference between the amounts of remaining tabs of different types |e - s|.

Examples

Input


4 2
1 1 -1 1


Output


2


Input


14 3
-1 1 -1 -1 1 -1 -1 1 -1 -1 1 -1 -1 1


Output


9

Note

In the first example we can choose b = 1 or b = 3. We will delete then one tab of each type and the remaining tabs are then all contain test information. Thus, e = 2 and s = 0 and |e - s| = 2.

In the second example, on the contrary, we can leave opened only tabs that have social networks opened in them.
#include <bits/stdc++.h>
using namespace std;
struct fast_ios {
  fast_ios() {
    cin.tie(nullptr), ios::sync_with_stdio(false),
        cout << fixed << setprecision(20);
  };
} fast_ios_;
template <class T>
auto add = [](T a, T b) -> T { return a + b; };
template <class T>
auto f_max = [](T a, T b) -> T { return max(a, b); };
template <class T>
auto f_min = [](T a, T b) -> T { return min(a, b); };
template <class T>
using V = vector<T>;
using Vl = V<long long int>;
using VVl = V<Vl>;
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (int i = 0; i < (int)v.size(); i++)
    os << v[i] << (i + 1 != v.size() ? " " : "");
  return os;
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
  for (T& in : v) is >> in;
  return is;
}
template <class T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int ceil(long long int a, long long int b) { return (a + b - 1) / b; }
long long int digit(long long int a) { return (long long int)log10(a); }
long long int e_dist(pair<long long int, long long int> a,
                     pair<long long int, long long int> b) {
  return abs(a.first - b.first) * abs(a.first - b.first) +
         abs(a.second - b.second) * abs(a.second - b.second);
}
long long int m_dist(pair<long long int, long long int> a,
                     pair<long long int, long long int> b) {
  return abs(a.first - b.first) + abs(a.second - b.second);
}
void Worshall_Floyd(VVl& g) {
  for (long long int k = (0), k_end_ = (((long long int)(g).size()));
       k < k_end_; ++k)
    for (long long int i = (0), i_end_ = (((long long int)(g).size()));
         i < i_end_; ++i)
      for (long long int j = (0), j_end_ = (((long long int)(g).size()));
           j < j_end_; ++j)
        chmin(g[i][j], g[i][k] + g[k][j]);
}
const long long int MOD1000000007 = 1000000007, MOD998244353 = 998244353,
                    INF = 1e18;
long long int dx[8] = {1, 0, -1, 0, 1, -1, 1, -1},
              dy[8] = {0, 1, 0, -1, -1, -1, 1, 1};
bool YN(bool flag) {
  cout << (flag ? "YES" : "NO") << '\n';
  return flag;
}
bool yn(bool flag) {
  cout << (flag ? "Yes" : "No") << '\n';
  return flag;
}
struct Edge {
  long long int from, to;
  long long int cost;
  Edge() {}
  Edge(long long int u, long long int v, long long int c) {
    cost = c;
    from = u;
    to = v;
  }
  bool operator<(const Edge& e) const { return cost < e.cost; }
};
struct WeightedEdge {
  long long int to;
  long long int cost;
  WeightedEdge(long long int v, long long int c = 1) {
    to = v;
    cost = c;
  }
  bool operator<(const WeightedEdge& e) const { return cost < e.cost; }
};
using WeightedGraph = V<V<WeightedEdge>>;
long long int N, K;
int main() {
  cin >> N >> K;
  Vl arr(N);
  cin >> arr;
  long long int curr = 0;
  for (long long int i = (0), i_end_ = (N); i < i_end_; ++i) curr += arr[i];
  long long int maxv = 0, _curr = curr;
  for (long long int i = (0), i_end_ = (K); i < i_end_; ++i) {
    _curr = curr;
    long long int idx = i;
    while (idx < N) {
      _curr -= arr[idx];
      idx += K;
    }
    chmax(maxv, abs(_curr));
  }
  cout << maxv << '\n';
}
