（寒假CF）惨淡一点

题解：类似于最大上升子序列

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

He writes n integers a₁, a₂, ..., 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 xmeans 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, ..., 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.

Sample Input

Input

5
1 0 0 1 0

Output

4

Input

4
1 0 0 1

Output

4

Hint

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<stdio.h>
int main()
{
    int n,i;
    int a[101];
    while(~scanf("%d",&n))
    {
        int sum=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        int k=0,max=0;
        for(i=0;i<n;i++)
        {
            if(a[i]==0)
            {
                k++;
            }
            else
            {
                if(k>0)
                k--;
            }
            if(k>=max)
            max=k;
        }
        if(max)
        printf("%d\n",max+sum);
        else
        printf("%d\n",max+sum-1);
    }
    return 0;
}