codeforces 327A

A. Flipping Game

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

第一行输入数字个数n，第二行输入n个0或者1，将第i个到第j个数替换为1-x，求替换后的1一共有多少个（1≤ i ≤ j ≤ n）。

开始的想法是怎么实现这个替换过程，实现起来很复杂，然后想了想是不是可以计算区域内0比1最多多几个，把这部分区域替换，然后将原来1的个数加上多的个数就是要求的结果。

#include<iostream>
using namespace std;
int main()
{
    int max = 0, n, i, sum = 0, a[105], p;
    cin >> n;
    p = 0;
    for (i = 0;i < n;i++)
    {
        cin >> a[i];
        sum += a[i];
    }
    if (sum == n)
    {
        cout << sum - 1 << endl;
        return 0;
    }
    for (i = 0;i < n;i++)
    {
        if (a[i] == 0)
            p++;
        else if (a[i] == 1 && p < 0)
            p = 0;
        else if (a[i] == 1 && p > 0)
            p--;
        if (max < p)
            max = p;
    }
    sum += max;
    cout << sum << endl;
    return 0;
}