#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  int arr[n], i, j = 0, num1 = 0, k = 0;
  int count[n + 1], c = 0;
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] == 1) c++;
  }
  if (c == n)
    cout << n - 1;
  else {
    int ans, ans2, pans = 0, l, sum = 0;
    for (i = 0; i < n; i++) {
      num1 = 0;
      for (j = i; j < n; j++) {
        if (arr[j] == 1) num1++;
        k = j - i + 1 - num1;
        l = c - num1;
        if (k > num1 && pans < k + l)
          pans = k + l;
        else if (pans < num1 + l)
          pans = num1 + l;
      }
    }
    cout << pans << endl;
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
  cin >> n;
  int arr[n], i, j = 0, num1 = 0, k = 0;
  int count[n + 1], c = 0;
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] == 1) c++;
  }
  if (c == n)
    cout << n - 1;
  else {
    int ans, ans2, pans = 0, l, sum = 0;
    for (i = 0; i < n; i++) {
      num1 = 0;
      for (j = i; j < n; j++) {
        if (arr[j] == 1) num1++;
        k = j - i + 1 - num1;
        l = c - num1;
        if (k > num1 && pans < k + l)
          pans = k + l;
        else if (pans < num1 + l)
          pans = num1 + l;
      }
    }
    cout << pans << endl;
  }
  return 0;
}

```