#include <bits/stdc++.h>
using namespace std;
int main() {
  int N, A[10000], count = 0, i, j, B[10000], Max[10000], ans;
  cin >> N;
  for (i = 0; i < N; i++) {
    cin >> A[i];
    if (A[i]) count++;
  }
  for (i = 0; i < N; i++) {
    if (i == 0 && A[i])
      B[i] = -1;
    else if (i == 0)
      B[i] = 1;
    else if (A[i])
      B[i] = B[i - 1] - 1;
    else
      B[i] = B[i - 1] + 1;
  }
  Max[N - 1] = B[N - 1];
  for (i = N - 2; i >= 0; i--) Max[i] = (B[i] > Max[i + 1]) ? B[i] : Max[i + 1];
  ans = 0;
  if (count == 0) ans = N;
  for (i = 0; i < N - 1; i++) {
    if (Max[i + 1] - B[i] > ans) ans = Max[i + 1] - B[i];
  }
  if (count != N)
    cout << count + ans;
  else {
    count--;
    cout << count;
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

Iahub got bored, so he invented a game to be played on paper. 

He writes n integers a1, a2, ..., an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1 ≤ i ≤ j ≤ n) and flips all values ak for which their positions are in range [i, j] (that is i ≤ k ≤ j). Flip the value of x means to apply operation x = 1 - x.

The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 100). In the second line of the input there are n integers: a1, a2, ..., an. It is guaranteed that each of those n values is either 0 or 1.

Output

Print an integer — the maximal number of 1s that can be obtained after exactly one move. 

Examples

Input

5
1 0 0 1 0


Output

4


Input

4
1 0 0 1


Output

4

Note

In the first case, flip the segment from 2 to 5 (i = 2, j = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].

In the second case, flipping only the second and the third element (i = 2, j = 3) will turn all numbers into 1.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int N, A[10000], count = 0, i, j, B[10000], Max[10000], ans;
  cin >> N;
  for (i = 0; i < N; i++) {
    cin >> A[i];
    if (A[i]) count++;
  }
  for (i = 0; i < N; i++) {
    if (i == 0 && A[i])
      B[i] = -1;
    else if (i == 0)
      B[i] = 1;
    else if (A[i])
      B[i] = B[i - 1] - 1;
    else
      B[i] = B[i - 1] + 1;
  }
  Max[N - 1] = B[N - 1];
  for (i = N - 2; i >= 0; i--) Max[i] = (B[i] > Max[i + 1]) ? B[i] : Max[i + 1];
  ans = 0;
  if (count == 0) ans = N;
  for (i = 0; i < N - 1; i++) {
    if (Max[i + 1] - B[i] > ans) ans = Max[i + 1] - B[i];
  }
  if (count != N)
    cout << count + ans;
  else {
    count--;
    cout << count;
  }
  return 0;
}

```