#include <bits/stdc++.h>
using namespace std;
int main() {
  int S, i = 0, one = 0, zero = 0, temp = 0;
  cin >> S;
  int ar[S];
  while (i < S) {
    cin >> ar[i];
    if (ar[i] == 1) {
      ar[i++] = -1;
      one++;
    } else
      ar[i++] = 1;
  }
  for (int k = 0; k < S; k++) {
    temp += ar[k];
    if (temp < 0) temp = 0;
    if (zero < temp) zero = temp;
  }
  if (zero)
    cout << one + zero << endl;
  else
    cout << --one << endl;
  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 S, i = 0, one = 0, zero = 0, temp = 0;
  cin >> S;
  int ar[S];
  while (i < S) {
    cin >> ar[i];
    if (ar[i] == 1) {
      ar[i++] = -1;
      one++;
    } else
      ar[i++] = 1;
  }
  for (int k = 0; k < S; k++) {
    temp += ar[k];
    if (temp < 0) temp = 0;
    if (zero < temp) zero = temp;
  }
  if (zero)
    cout << one + zero << endl;
  else
    cout << --one << endl;
  return 0;
}

```