A - A Flipping Game

这道题判断如何选择区间进行01变换让数列中的1个数最多，可以用暴力做法来做，每选择一个区间求出一个值，最后找到一个最大值。

Iahub got bored, so he invented a game to be played on paper.

He writes n integers a ₁, a ₂, ..., a _n. 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 a _k 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: a ₁, a ₂, ..., a _n. 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.

#include <string>
#include <iostream>
using namespace std;
int f[101],g[101];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    scanf("%d",&f[i]);
    int sum=0,m=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=i;j<=n;j++)
        {
            sum=0;
            for(int s=1;s<=n;s++)//1到n的和
                sum+=f[s];
            for(int k=i;k<=j;k++)//取每一段相减，取反相加
            {
                sum=sum-f[k]+1-f[k];
            }
            m=max(m,sum);//最大值
        }
    }
    printf("%d\n",m);
}