#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, a[105], sum[105], ans = INT_MIN;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sum[0] = a[0];
  for (int i = 1; i < n; i++) sum[i] = sum[i - 1] + a[i];
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      int x;
      int inc;
      if (i != 0) {
        x = (j - i + 1) - (sum[j] - sum[i - 1]);
        inc = x - (sum[j] - sum[i - 1]);
      } else {
        x = (j - i + 1) - sum[j];
        inc = x - sum[j];
      }
      int y;
      y = sum[n - 1] + inc;
      ans = max(ans, y);
    }
  }
  cout << ans << 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, a[105], sum[105], ans = INT_MIN;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sum[0] = a[0];
  for (int i = 1; i < n; i++) sum[i] = sum[i - 1] + a[i];
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      int x;
      int inc;
      if (i != 0) {
        x = (j - i + 1) - (sum[j] - sum[i - 1]);
        inc = x - (sum[j] - sum[i - 1]);
      } else {
        x = (j - i + 1) - sum[j];
        inc = x - sum[j];
      }
      int y;
      y = sum[n - 1] + inc;
      ans = max(ans, y);
    }
  }
  cout << ans << endl;
  return 0;
}

```