#include <bits/stdc++.h>
int main() {
  int n, i, p = 0, q = 0, k = 0, j, max = 0;
  scanf("%d", &n);
  int a[n], b[n], c[n];
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 0) {
      b[q++] = k;
      c[p++] = i;
    } else {
      k++;
    }
  }
  for (i = 0; i < p; i++) {
    for (j = i + 1; j < p; j++) {
      if (k + c[j] - c[i] + 1 - 2 * (b[j] - b[i]) > max) {
        max = k + c[j] - c[i] + 1 - 2 * (b[j] - b[i]);
      }
    }
  }
  if (n == 1 && a[0] == 0)
    printf("1");
  else if (n > 1 && p == 0)
    printf("%d", n - 1);
  else
    printf("%d", max);
}

### Prompt

Please create a solution in CPP 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>
int main() {
  int n, i, p = 0, q = 0, k = 0, j, max = 0;
  scanf("%d", &n);
  int a[n], b[n], c[n];
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 0) {
      b[q++] = k;
      c[p++] = i;
    } else {
      k++;
    }
  }
  for (i = 0; i < p; i++) {
    for (j = i + 1; j < p; j++) {
      if (k + c[j] - c[i] + 1 - 2 * (b[j] - b[i]) > max) {
        max = k + c[j] - c[i] + 1 - 2 * (b[j] - b[i]);
      }
    }
  }
  if (n == 1 && a[0] == 0)
    printf("1");
  else if (n > 1 && p == 0)
    printf("%d", n - 1);
  else
    printf("%d", max);
}

```