#include <bits/stdc++.h>
using namespace std;
long long r = -1, c, x, n, t, y, counb, couna, m, b, z = 10e18, a, summin,
          summax, e, l, k;
long long mod = 1e9 + 7, maxx = 0, minn = 10e18;
long long arr[1000001];
long long dif[1000001], ar[4000];
char cv;
string str, air, er;
bool s = 0, w = 0;
int main() {
  long long res = 0;
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    c += a;
    x += 1 - a * 2;
    r = max(r, x - m);
    m = min(x, m);
  }
  cout << r + c;
}

### 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;
long long r = -1, c, x, n, t, y, counb, couna, m, b, z = 10e18, a, summin,
          summax, e, l, k;
long long mod = 1e9 + 7, maxx = 0, minn = 10e18;
long long arr[1000001];
long long dif[1000001], ar[4000];
char cv;
string str, air, er;
bool s = 0, w = 0;
int main() {
  long long res = 0;
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    c += a;
    x += 1 - a * 2;
    r = max(r, x - m);
    m = min(x, m);
  }
  cout << r + c;
}

```