There are N squares aligned in a row. The i-th square from the left contains an integer a_i.

Initially, all the squares are white. Snuke will perform the following operation some number of times:

* Select K consecutive squares. Then, paint all of them white, or paint all of them black. Here, the colors of the squares are overwritten.



After Snuke finishes performing the operation, the score will be calculated as the sum of the integers contained in the black squares. Find the maximum possible score.

Constraints

* 1≤N≤10^5
* 1≤K≤N
* a_i is an integer.
* |a_i|≤10^9

Input

The input is given from Standard Input in the following format:


N K
a_1 a_2 ... a_N


Output

Print the maximum possible score.

Examples

Input

5 3
-10 10 -10 10 -10


Output

10


Input

4 2
10 -10 -10 10


Output

20


Input

1 1
-10


Output

0


Input

10 5
5 -4 -5 -8 -4 7 2 -4 0 7


Output

17
#include <bits/stdc++.h>
using namespace std;
double EPS = 1e-9;
int INF = 1000000005;
long long INFF = 1000000000000000005LL;
double PI = acos(-1);
int dirx[8] = {-1, 0, 0, 1, -1, -1, 1, 1};
int diry[8] = {0, 1, -1, 0, -1, 1, -1, 1};
inline string IntToString(long long a) {
  char x[100];
  sprintf(x, "%lld", a);
  string s = x;
  return s;
}
inline long long StringToInt(string a) {
  char x[100];
  long long res;
  strcpy(x, a.c_str());
  sscanf(x, "%lld", &res);
  return res;
}
inline string GetString(void) {
  char x[1000005];
  scanf("%s", x);
  string s = x;
  return s;
}
inline string uppercase(string s) {
  int n = (int)s.size();
  for (int(i) = (0); (i) < (n); ++(i))
    if (s[i] >= 'a' && s[i] <= 'z') s[i] = s[i] - 'a' + 'A';
  return s;
}
inline string lowercase(string s) {
  int n = (int)s.size();
  for (int(i) = (0); (i) < (n); ++(i))
    if (s[i] >= 'A' && s[i] <= 'Z') s[i] = s[i] - 'A' + 'a';
  return s;
}
inline void OPEN(string s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
class SegmentTree {
 public:
  long long prop[800005];
  long long val[800005];
  void build() {
    memset(prop, 0, sizeof(prop));
    memset(val, 0, sizeof(val));
  }
  void push(int ix, int L, int R) {
    val[ix * 2 + 1] += prop[ix];
    val[ix * 2 + 2] += prop[ix];
    prop[ix * 2 + 1] += prop[ix];
    prop[ix * 2 + 2] += prop[ix];
    prop[ix] = 0;
  }
  void update(int ix, int L, int R, int x, int y, long long v) {
    if (x <= L && R <= y) {
      val[ix] += v;
      prop[ix] += v;
      return;
    }
    if (R < x || y < L) return;
    push(ix, L, R);
    int M = (L + R) >> 1;
    update(ix * 2 + 1, L, M, x, y, v);
    update(ix * 2 + 2, M + 1, R, x, y, v);
    val[ix] = max(val[ix * 2 + 1], val[ix * 2 + 2]);
  }
  long long query(int ix, int L, int R, int x, int y) {
    if (x <= L && R <= y) return val[ix];
    if (R < x || y < L) return -INFF;
    push(ix, L, R);
    int M = (L + R) >> 1;
    return max(query(ix * 2 + 1, L, M, x, y),
               query(ix * 2 + 2, M + 1, R, x, y));
  }
};
int n, k;
int data[100005];
long long sum[100005];
int last = -INF;
long long risan = 0;
long long dp[100005];
SegmentTree hitam, putih;
long long jum(int a, int b) { return (a == 0 ? sum[b] : sum[b] - sum[a - 1]); }
int main() {
  putih.build();
  hitam.build();
  scanf("%d %d", &n, &k);
  for (int(i) = (0); (i) < (n); ++(i)) {
    scanf("%d", &data[i]);
    sum[i] = (i == 0 ? 0 : sum[i - 1]);
    sum[i] += data[i];
  }
  if (k == 1) {
    for (int(i) = (0); (i) < (n); ++(i)) risan += (data[i] >= 0 ? data[i] : 0);
    printf("%lld\n", risan);
    return 0;
  }
  for (int(i) = (0); (i) < (n); ++(i)) risan += data[i];
  for (int(i) = (n - k + 1); (i) < (n); ++(i)) {
    hitam.update(0, 0, n, i, i, -INFF);
  }
  for (int(i) = (n - k); (i) >= (0); --(i)) {
    putih.update(0, 0, n, i + 1, i + k - 1, -data[i]);
    long long res1 = -hitam.query(0, 0, n, i + 1, i + k - 1);
    long long res2 = -putih.query(0, 0, n, i + 1, i + k - 1);
    dp[i] = min(res1, res2);
    hitam.update(0, 0, n, i, i, -dp[i]);
    putih.update(0, 0, n, i, i, -dp[i]);
  }
  risan -= dp[0];
  putih.build();
  hitam.build();
  for (int(i) = (n - k + 1); (i) < (n); ++(i)) {
    hitam.update(0, 0, n, i, i, -INFF);
  }
  for (int(i) = (n - k); (i) >= (0); --(i)) {
    hitam.update(0, 0, n, i + 1, i + k - 1, data[i]);
    long long res1 = hitam.query(0, 0, n, i + 1, i + k - 1);
    long long res2 = putih.query(0, 0, n, i + 1, i + k - 1);
    dp[i] = max(res1, res2);
    hitam.update(0, 0, n, i, i, dp[i]);
    putih.update(0, 0, n, i, i, dp[i]);
  }
  printf("%lld\n", max(risan, dp[0]));
  return 0;
}
