#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
const long long INF = 1LL << 60;
int n, k;
long long a[maxn], e[maxn];
long long s[maxn];
long long f[maxn];
long long min_pre[maxn];
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> k;
  long long sum = 0, min_e = INF;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    sum += a[i];
  }
  for (int i = n; i >= 1; --i) {
    s[i] = a[i] + s[i + 1];
  }
  min_pre[0] = INF;
  for (int i = 1; i <= n; ++i) {
    cin >> e[i];
    min_pre[i] = min(min_pre[i - 1], e[i]);
    if (i != n) min_e = min(min_e, e[i]);
  }
  for (int i = n; i >= 1; --i) {
    f[i] = max(f[i + 1], s[i] - e[i]);
  }
  if (k >= 2) {
    cout << max(0LL, max(sum - min_e, a[n] - e[n])) << endl;
    return 0;
  } else if (k == 0) {
    cout << f[1] << endl;
  } else if (k == 1) {
    long long ans = 0;
    for (int i = 1; i < n; ++i) {
      ans = max(ans, s[1] - s[i + 1] - min_pre[i] + f[i + 1]);
    }
    long long t = 0;
    for (int i = 1; i <= n - 2; ++i) {
      t = max(t, s[i] - e[i]);
      ans = max(ans, t - a[i + 1] + max(a[i + 1] - e[i + 1], 0LL));
      ans = max(ans, f[i + 1]);
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

Mr. Chanek is currently participating in a science fair that is popular in town. He finds an exciting puzzle in the fair and wants to solve it.

There are N atoms numbered from 1 to N. These atoms are especially quirky. Initially, each atom is in normal state. Each atom can be in an excited. Exciting atom i requires D_i energy. When atom i is excited, it will give A_i energy. You can excite any number of atoms (including zero).

These atoms also form a peculiar one-way bond. For each i, (1 ≤ i < N), if atom i is excited, atom E_i will also be excited at no cost. Initially, E_i = i+1. Note that atom N cannot form a bond to any atom.

Mr. Chanek must change exactly K bonds. Exactly K times, Mr. Chanek chooses an atom i, (1 ≤ i < N) and changes E_i to a different value other than i and the current E_i. Note that an atom's bond can remain unchanged or changed more than once. Help Mr. Chanek determine the maximum energy that he can achieve!

note: You must first change exactly K bonds before you can start exciting atoms.

Input

The first line contains two integers N K (4 ≤ N ≤ 10^5, 0 ≤ K < N), the number of atoms, and the number of bonds that must be changed.

The second line contains N integers A_i (1 ≤ A_i ≤ 10^6), which denotes the energy given by atom i when on excited state.

The third line contains N integers D_i (1 ≤ D_i ≤ 10^6), which denotes the energy needed to excite atom i.

Output

A line with an integer that denotes the maximum number of energy that Mr. Chanek can get.

Example

Input


6 1
5 6 7 8 10 2
3 5 6 7 1 10


Output


35

Note

An optimal solution to change E_5 to 1 and then excite atom 5 with energy 1. It will cause atoms 1, 2, 3, 4, 5 be excited. The total energy gained by Mr. Chanek is (5 + 6 + 7 + 8 + 10) - 1 = 35.

Another possible way is to change E_3 to 1 and then exciting atom 3 (which will excite atom 1, 2, 3) and exciting atom 4 (which will excite atom 4, 5, 6). The total energy gained by Mr. Chanek is (5 + 6 + 7 + 8 + 10 + 2) - (6 + 7) = 25 which is not optimal.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
const long long INF = 1LL << 60;
int n, k;
long long a[maxn], e[maxn];
long long s[maxn];
long long f[maxn];
long long min_pre[maxn];
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> k;
  long long sum = 0, min_e = INF;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    sum += a[i];
  }
  for (int i = n; i >= 1; --i) {
    s[i] = a[i] + s[i + 1];
  }
  min_pre[0] = INF;
  for (int i = 1; i <= n; ++i) {
    cin >> e[i];
    min_pre[i] = min(min_pre[i - 1], e[i]);
    if (i != n) min_e = min(min_e, e[i]);
  }
  for (int i = n; i >= 1; --i) {
    f[i] = max(f[i + 1], s[i] - e[i]);
  }
  if (k >= 2) {
    cout << max(0LL, max(sum - min_e, a[n] - e[n])) << endl;
    return 0;
  } else if (k == 0) {
    cout << f[1] << endl;
  } else if (k == 1) {
    long long ans = 0;
    for (int i = 1; i < n; ++i) {
      ans = max(ans, s[1] - s[i + 1] - min_pre[i] + f[i + 1]);
    }
    long long t = 0;
    for (int i = 1; i <= n - 2; ++i) {
      t = max(t, s[i] - e[i]);
      ans = max(ans, t - a[i + 1] + max(a[i + 1] - e[i + 1], 0LL));
      ans = max(ans, f[i + 1]);
    }
    cout << ans << endl;
  }
  return 0;
}

```