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
/*
Author: QAQ-Automaton
LANG: C++
PROG: b.cpp
Mail: cnyalilk@vip.qq.com
*/
#include<bits/stdc++.h>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %lld\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %lld\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pii;
const signed inf=0x3f3f3f3f;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T>ll chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>ll chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>ll dcmp(T a,T b){return a>b;}
template<ll *a>ll cmp_a(ll x,ll y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
namespace io {
	const ll SIZE = (1 << 21) + 1;
	char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; ll f, qr;
	// getchar
	#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
	// prll the remaining part
	inline void flush () {
		fwrite (obuf, 1, oS - obuf, stdout);
		oS = obuf;
	}
	// putchar
	inline void putc (char x) {
		*oS ++ = x;
		if (oS == oT) flush ();
	}
	// input a signed lleger
	inline bool read (signed &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
		return 1;
	}

	inline bool read (long long &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;else if(c==EOF)return 0;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
		return 1;
	}
	inline bool read (char &x) {
		x=gc();
		return x!=EOF;
	}
	inline bool read(char *x){
		while((*x=gc())=='\n' || *x==' '||*x=='\r')if(*x==EOF)return 0;
		while(!(*x=='\n'||*x==' '||*x=='\r'))*(++x)=gc();
		*x=0;
		return 1;
	}
	template<typename A,typename ...B>
	inline bool read(A &x,B &...y){
		return read(x)&&read(y...);
	}
	// prll a signed lleger
	inline bool write (signed x) {
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
		while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
		while (qr) putc (qu[qr --]);
		return 0;
	}

	inline bool write (long long x) {
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
		while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
		while (qr) putc (qu[qr --]);
		return 0;
	}
	inline bool write (char x) {
		putc(x);
		return 0;
	}
	inline bool write(const char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	inline bool write(char *x){
		while(*x){putc(*x);++x;}
		return 0;
	}
	template<typename A,typename ...B>
	inline bool write(A x,B ...y){
		return write(x)||write(y...);
	}
	//no need to call flush at the end manually!
	struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
ll a[100005],b[100005];
int main(){
#ifdef QAQAutoMaton 
	freopen("b.in","r",stdin);
	freopen("b.out","w",stdout);
#endif
	ll n,k,m;
	read(n,k);
	m=n-k;
	for(ll i=1;i<=n;++i)read(a[i]);
	for(ll i=1;i<=n;++i){
		b[i]=(a[i]>0?a[i]:0)+b[i-1];
		a[i]+=a[i-1];
	}
	ll ans=0;
	for(ll i=0;i<=m;++i){
		chkmax(ans,max(a[i+k]-a[i],0)+b[i]+b[n]-b[i+k]);
	}
	write(ans,'\n');
	return 0;
}

