luogu2034

/*
 * 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 + 10;

LL tot, d, n, k;
LL p[N], head = 1, tail = 1;
LL q[N], f[N], ans;

int main() {
      std:: cin >> n >> k; 
    for(int i = 1; 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);
}