#include <bits/stdc++.h>
using namespace std;
mt19937_64 rang(
    chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim) {
  uniform_int_distribution<int> uid(0, lim - 1);
  return uid(rang);
}
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int kadanes(vector<int> &v, const int n) {
  int mx = *max_element(v.begin(), v.end());
  if (mx <= 0) return mx;
  int mx_so_far = 0;
  int mx_end_here = 0;
  for (int i = 0; i < n; i++) {
    mx_end_here += v[i];
    mx_end_here = max(0, mx_end_here);
    mx_so_far = max(mx_so_far, mx_end_here);
  }
  return mx_so_far;
}
void solve() {
  int n;
  cin >> n;
  vector<int> v(n);
  int ones = 0;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    if (v[i]) ones++;
  }
  vector<int> t(n);
  for (int i = 0; i < n; i++) {
    t[i] = (v[i] == 0) ? 1 : -1;
  }
  int net = kadanes(t, n);
  cout << ones + net << "\n";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### 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>
using namespace std;
mt19937_64 rang(
    chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim) {
  uniform_int_distribution<int> uid(0, lim - 1);
  return uid(rang);
}
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int kadanes(vector<int> &v, const int n) {
  int mx = *max_element(v.begin(), v.end());
  if (mx <= 0) return mx;
  int mx_so_far = 0;
  int mx_end_here = 0;
  for (int i = 0; i < n; i++) {
    mx_end_here += v[i];
    mx_end_here = max(0, mx_end_here);
    mx_so_far = max(mx_so_far, mx_end_here);
  }
  return mx_so_far;
}
void solve() {
  int n;
  cin >> n;
  vector<int> v(n);
  int ones = 0;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    if (v[i]) ones++;
  }
  vector<int> t(n);
  for (int i = 0; i < n; i++) {
    t[i] = (v[i] == 0) ? 1 : -1;
  }
  int net = kadanes(t, n);
  cout << ones + net << "\n";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```