#include <bits/stdc++.h>
using namespace std;
int MaxSumSub(vector<int> &S, int sum, int n) {
  if (sum == n) return n - 1;
  int sum2, ans;
  sum2 = 0;
  ans = 0;
  for (int k = 0; k < n; ++k) {
    if (sum2 + S[k] > 0)
      sum2 += S[k];
    else
      sum2 = 0;
    ans = max(ans, sum2);
  }
  return ans + sum;
}
int main() {
  int n, tmp, sum = 0;
  vector<int> num;
  cin >> n;
  while (n--) {
    cin >> tmp;
    if (tmp == 1) {
      sum++;
      tmp = -1;
    } else
      tmp = 1;
    num.push_back(tmp);
  }
  cout << MaxSumSub(num, sum, num.size());
  return 0;
}

### Prompt

Your task is to create 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;
int MaxSumSub(vector<int> &S, int sum, int n) {
  if (sum == n) return n - 1;
  int sum2, ans;
  sum2 = 0;
  ans = 0;
  for (int k = 0; k < n; ++k) {
    if (sum2 + S[k] > 0)
      sum2 += S[k];
    else
      sum2 = 0;
    ans = max(ans, sum2);
  }
  return ans + sum;
}
int main() {
  int n, tmp, sum = 0;
  vector<int> num;
  cin >> n;
  while (n--) {
    cin >> tmp;
    if (tmp == 1) {
      sum++;
      tmp = -1;
    } else
      tmp = 1;
    num.push_back(tmp);
  }
  cout << MaxSumSub(num, sum, num.size());
  return 0;
}

```