#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline const T MAX(T a, T b) {
  return ((a) > (b) ? (a) : (b));
}
template <typename T>
inline const T MIN(T a, T b) {
  return ((a) > (b) ? (b) : (a));
}
template <typename T>
inline const T ABS(T a) {
  return (((a) > 0) ? (a) : (-(a)));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> A = vector<int>(n);
  for (int i = 0; i < (int)n; ++i) cin >> A[i];
  int max_count = -1e9;
  for (int i = 0; i < n; ++i) {
    for (int j = i; j < n; ++j) {
      int cnt = 0;
      for (int k = 0; k < n; ++k) {
        if (i <= k && k <= j)
          cnt += (A[k] ^ 1);
        else
          cnt += A[k];
      }
      max_count = MAX(max_count, cnt);
    }
  }
  cout << max_count << endl;
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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;
template <typename T>
inline const T MAX(T a, T b) {
  return ((a) > (b) ? (a) : (b));
}
template <typename T>
inline const T MIN(T a, T b) {
  return ((a) > (b) ? (b) : (a));
}
template <typename T>
inline const T ABS(T a) {
  return (((a) > 0) ? (a) : (-(a)));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> A = vector<int>(n);
  for (int i = 0; i < (int)n; ++i) cin >> A[i];
  int max_count = -1e9;
  for (int i = 0; i < n; ++i) {
    for (int j = i; j < n; ++j) {
      int cnt = 0;
      for (int k = 0; k < n; ++k) {
        if (i <= k && k <= j)
          cnt += (A[k] ^ 1);
        else
          cnt += A[k];
      }
      max_count = MAX(max_count, cnt);
    }
  }
  cout << max_count << endl;
  return 0;
}

```