/*

 * 正难则反

 * f[i] 表示前 i 个数中被删除的数的最小和

 * f[i] = min(f[j]) + num, i - k + 1 <= j < i;

 * 单调队列维护

 */

#include <bits/stdc++.h>

#define LL long long

const int N = 1e5 + ;

LL tot, d, n, k;

LL p[N], head = , tail = ;

LL q[N], f[N], ans;

int main() {

      std:: cin >> n >> k;

    for(int i = ; i <= n; ++ i)

    {

        std:: cin >> d;

        tot += d;

        f[i] = q[head] + 1LL * d;

        while(head <= tail && q[tail] >= f[i]) tail --;

        q[++ tail] = f[i], p[tail] = i;

        while(head <= tail && p[head] < i - k) head ++;

    }

    for(int i = n - k; i <= n; ++ i) ans = std:: max(ans, 1LL * tot - 1LL * f[i]);

    printf("%lld", ans);

}