Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

5 3
-10 10 -10 10 -10

Sample Output 1

10

Paint the following squares black: the second, third and fourth squares from the left.

Sample Input 2

4 2
10 -10 -10 10

Sample Output 2

20

One possible way to obtain the maximum score is as follows:

Paint the following squares black: the first and second squares from the left.

Paint the following squares black: the third and fourth squares from the left.

Paint the following squares white: the second and third squares from the left.

Sample Input 3

1 1
-10

Sample Output 3

0

Sample Input 4

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

Sample Output 4

17

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

#define FOR(i,a,b)  for(ll (i)=a;(i)<(b);++(i))
#define RFOR(i,a,b) for(ll (i)=a;(i)>=(b);--(i))
#define REP(i,n)    FOR(i,0,n)
#define RREP(i,n)   RFOR(i,n,0)
#define ALL(v)      (v).begin(),(v).end()
#define BIT(n)      (1LL<<(n))
#define IN(x,l,r)   ((l)<=(x)&&(x)<=(r))
#define LEN(s)      (string(s)).length()
#define NPOS        string::npos
#define SHOW(x)     cout<<#x<<" = "<<x<<endl
#define SORT(v)     sort(ALL(v))
#define SZ(v)       (ll)(v).size()
#define UNIQ(v)     SORT(v);(v).erase(unique(ALL(v)),(v).end())
#define PERM(v)     for(bool p=1;p;p=next_permutation(ALL(v)))
#define PQ(T)       priority_queue<T,vector<T>,greater<T>>

template<class T>string join(const vector<T>& v,string d=" ")
{stringstream s;REP(i,SZ(v))s<<d<<v[i];return s.str().substr(LEN(d));}

template<class T>ostream& operator<<(ostream& s,const vector<T>& v)
{s<<"{";if(SZ(v))s<<join(v,", ");s<<"}";return s;}

template<class T>bool contain(const string& s,const T& v)
{return s.find(v)!=NPOS;}

template<class T,class U>string& replace(string& s,const T& f,const U& t)
{for(auto p=s.find(f);p!=NPOS;p=s.find(f,p+LEN(t)))s.replace(p,LEN(f),t);}

const ll INF = 1e15;
const ll MOD = 1e9+7;
const double EPS = 1e-10;

int dy[] = {0, 0, 1, -1};
int dx[] = {1, -1, 0, 0};

int main() {
    ll N, K; cin >> N >> K;
    vector<ll> a(N); REP(i, N) cin >> a[i];
    
    vector<ll> sum(N+1, 0);
    REP(i, N) sum[i+1] = sum[i] + a[i];
    
    vector<ll> pls(N+1, 0);
    REP(i, N) pls[i+1] = pls[i] + (a[i] > 0 ? a[i] : 0);
    
    ll ans = 0;
    FOR(i, 1, N-K+2) {
        ll b = sum[i+K-1]-sum[i-1];
        if (b < 0) b = 0;
        b += pls[i-1];
        b += pls[N] - pls[i+K-1];
        ans = max(ans, b);
    }
    cout << ans << endl;
    
    return 0;
}