#include <bits/stdc++.h>
using namespace std;
long long n, k;
long long a[100005];
long long d[100005];
long long z[100005];
long long prefmin[100005];
long long suffmina[100005];
int main() {
  ios::sync_with_stdio(!cin.tie(0));
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> d[i];
  }
  z[n] = a[n];
  for (int i = n - 1; i >= 1; i--) {
    z[i] = z[i + 1] + a[i];
  }
  long long sol = 0;
  if (k == 0) {
    for (int i = 1; i <= n; i++) {
      sol = max(sol, z[i] - d[i]);
    }
  } else if (k >= 2) {
    for (int i = 1; i < n; i++) {
      sol = max(sol, z[1] - d[i]);
    }
    sol = max(sol, a[n] - d[n]);
  } else {
    prefmin[1] = d[1];
    for (int i = 2; i <= n; i++) {
      prefmin[i] = min(prefmin[i - 1], d[i]);
    }
    suffmina[n - 1] = a[n - 1];
    for (int i = n - 2; i >= 1; i--) {
      suffmina[i] = min(suffmina[i + 1], a[i]);
    }
    sol = max(sol, a[n] - d[n]);
    for (int i = 1; i < n; i++) {
      sol = max(sol, z[1] - a[n] - d[i] + max(0ll, a[n] - d[n]));
    }
    for (int i = 2; i < n; i++) {
      long long v = z[i] - d[i];
      v += max(0ll, z[1] - z[i] - prefmin[i - 1]);
      sol = max(sol, v);
    }
    for (int i = 1; i < n - 1; i++) {
      sol = max(sol, z[i] - d[i] - suffmina[i + 1]);
    }
  }
  cout << sol << '\n';
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
long long n, k;
long long a[100005];
long long d[100005];
long long z[100005];
long long prefmin[100005];
long long suffmina[100005];
int main() {
  ios::sync_with_stdio(!cin.tie(0));
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> d[i];
  }
  z[n] = a[n];
  for (int i = n - 1; i >= 1; i--) {
    z[i] = z[i + 1] + a[i];
  }
  long long sol = 0;
  if (k == 0) {
    for (int i = 1; i <= n; i++) {
      sol = max(sol, z[i] - d[i]);
    }
  } else if (k >= 2) {
    for (int i = 1; i < n; i++) {
      sol = max(sol, z[1] - d[i]);
    }
    sol = max(sol, a[n] - d[n]);
  } else {
    prefmin[1] = d[1];
    for (int i = 2; i <= n; i++) {
      prefmin[i] = min(prefmin[i - 1], d[i]);
    }
    suffmina[n - 1] = a[n - 1];
    for (int i = n - 2; i >= 1; i--) {
      suffmina[i] = min(suffmina[i + 1], a[i]);
    }
    sol = max(sol, a[n] - d[n]);
    for (int i = 1; i < n; i++) {
      sol = max(sol, z[1] - a[n] - d[i] + max(0ll, a[n] - d[n]));
    }
    for (int i = 2; i < n; i++) {
      long long v = z[i] - d[i];
      v += max(0ll, z[1] - z[i] - prefmin[i - 1]);
      sol = max(sol, v);
    }
    for (int i = 1; i < n - 1; i++) {
      sol = max(sol, z[i] - d[i] - suffmina[i + 1]);
    }
  }
  cout << sol << '\n';
}

```