#include <bits/stdc++.h>
using namespace std;
const int Max = 1e5 + 5;
const int Mod = 998244353;
const int inf = 0x3f3f3f3f;
long long Demand[Max], worth[Max];
long long suf[Max], Suff[Max];
int main(void) {
  std::ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> worth[i];
  for (int i = 1; i <= n; i++) cin >> Demand[i];
  long long MinDemand = inf;
  for (int i = n; i > 0; i--) {
    suf[i] = suf[i + 1] + worth[i];
    Suff[i] = max(Suff[i + 1], suf[i] - Demand[i]);
    if (i != n) MinDemand = min(MinDemand, Demand[i]);
  }
  if (k == 0)
    cout << Suff[1] << '\n';
  else if (k >= 2) {
    cout << max({0ll, worth[n] - Demand[n], suf[1] - MinDemand}) << '\n';
  } else {
    long long persum = worth[1], mincost = Demand[1];
    long long ans = 0;
    for (int i = 2; i < n; i++) {
      mincost = min(mincost, Demand[i]);
      persum += worth[i];
      ans = max(ans, persum - mincost + Suff[i + 1]);
      ans = max(ans,
                suf[1] - Demand[1] - worth[i] + max(0ll, worth[i] - Demand[i]));
      ans = max(ans, Suff[i]);
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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 Max = 1e5 + 5;
const int Mod = 998244353;
const int inf = 0x3f3f3f3f;
long long Demand[Max], worth[Max];
long long suf[Max], Suff[Max];
int main(void) {
  std::ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> worth[i];
  for (int i = 1; i <= n; i++) cin >> Demand[i];
  long long MinDemand = inf;
  for (int i = n; i > 0; i--) {
    suf[i] = suf[i + 1] + worth[i];
    Suff[i] = max(Suff[i + 1], suf[i] - Demand[i]);
    if (i != n) MinDemand = min(MinDemand, Demand[i]);
  }
  if (k == 0)
    cout << Suff[1] << '\n';
  else if (k >= 2) {
    cout << max({0ll, worth[n] - Demand[n], suf[1] - MinDemand}) << '\n';
  } else {
    long long persum = worth[1], mincost = Demand[1];
    long long ans = 0;
    for (int i = 2; i < n; i++) {
      mincost = min(mincost, Demand[i]);
      persum += worth[i];
      ans = max(ans, persum - mincost + Suff[i + 1]);
      ans = max(ans,
                suf[1] - Demand[1] - worth[i] + max(0ll, worth[i] - Demand[i]));
      ans = max(ans, Suff[i]);
    }
    cout << ans << '\n';
  }
  return 0;
}

```