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>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define rng(a) a.begin(),a.end()
#define maxs(x,y) (x = max(x,y))
#define mins(x,y) (x = min(x,y))
#define limit(x,l,r) max(l,min(x,r))
#define lims(x,l,r) (x = max(l,min(x,r)))
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define pb push_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcountll
#define uni(x) x.erase(unique(rng(x)),x.end())
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,v(T),greater<T> >
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define newline puts("")
#define v(T) vector<T>
#define vv(T) v(v(T))
using namespace std;
typedef long long int ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
inline int in() { int x; scanf("%d",&x); return x;}
template<typename T>inline istream& operator>>(istream&i,v(T)&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(const v(T)&v)
{stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>inline ostream& operator<<(ostream&o,const v(T)&v)
{if(sz(v))o<<join(v);return o;}
template<typename T1,typename T2>inline istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<typename T1,typename T2>inline ostream& operator<<(ostream&o,const pair<T1,T2>&v)
{return o<<v.fi<<","<<v.se;}
template<typename T>inline ll suma(const v(T)& a) { ll res(0); for (auto&& x : a) res += x; return res;}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
#define dame { puts("-1"); return 0;}
#define yn {puts("Yes");}else{puts("No");}
const int MX = 200005;
#define RET(ans) {cout<<ans<<endl;return 0;} 

    // 二次元ベクターの基本
    /*
    vector<vector<int>> dp; // 宣言
    dp.resize(n); // 1次元めの要素数決定
    dp[i].push_back(int); // プッシュバック
    rep(i,n){
        sort(dp[i].begin(),dp[i].end()); // 二次元めを昇順ソート
    }
    */

    // 整数スキャン（複数）
    /*
    int x,y,z;
    scanf("%d%d%d",&x,&y,&z);

    // n個の整数のスキャン
    /*
    ll a[n] = {};
    rep(i,n){
        scanf("%lld",&a[i]);
    }
    */

    // 文字列スキャン
    /*
    string s; cin >> s;
    */

    // n個の文字列スキャン
    /*
    vector<string> slist;
    rep(i,n){
        string s; cin >> s;
        slist.push_back(s);
    }
    */

int main() {
    int n,k;
    scanf("%d%d",&n,&k);
    ll a[n] = {};
    ll ans = 0;
    rep(i,n){
        scanf("%lld",&a[i]);
        if(a[i]>0){
            ans += a[i];
        }
    }
    ll b[n][2];
    rep(i,n){
        b[i][0]=0;
        b[i][1]=0;
    }

    rep(i,k){
        if(a[i]>0){
            b[0][0] += a[i];
        }else{
            b[0][1] += -a[i];
        }
    }

    srep(i,1,n+1-k){
        b[i][0] = b[i-1][0];
        b[i][1] = b[i-1][1];
        if(a[i-1]>0){
            b[i][0] -= a[i-1]
        }else{
            b[i][1] -= -a[i-1];
        }
        if(a[k-1+i]>0){
            b[i][0] += a[k-1+i];
        }else{
            b[i][1] += -a[k-1+i];
        }
    }
    ll mi = b[0][0];
    rep(i,n+1-k){
        mi = min(mi,b[i][0]);
        mi = min(mi,b[i][1]);
    }
    
    
    cout << ans - mi << endl;
    return 0;
}
 
 
