Codeforces Round #442 (Div. 2) B

Topic links: http://codeforces.com/contest/877/problem/B

B. Nikita and string

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

One day Nikita found the string containing letters "a" and "b" only.

Nikita thinks that string is beautiful if it can be cut into 3 strings (possibly empty) without changing the order of the letters, where the 1-st and the 3-rd one contain only letters "a" and the 2-nd contains only letters "b".

Nikita wants to make the string beautiful by removing some (possibly none) of its characters, but without changing their order. What is the maximum length of the string he can get?

Input

The first line contains a non-empty string of length not greater than 5 000 containing only lowercase English letters "a" and "b".

Output

Print a single integer — the maximum possible size of beautiful string Nikita can get.

Examples

input      abba

output    4

input       bab

output     2

Note

It the first sample the string is already beautiful.

In the second sample he needs to delete one of "b" to make it beautiful.

The meaning of problems: Given a string containing only ab, to the character string is divided into 3 sections and the rear section of the front section comprising only a, B contains only an intermediate, can be used instead of spaces after a delete operation, the longest asked string

Ideas: record a, b character prefix and then cut-off point to enumerate

code show as below:

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int suma[5005];
int sumb[5005];
char str[5005];
int main()
{
    scanf("%s",str+1);
    int ans=strlen(str+1);
    for(int i=1;i<=ans;i++)
    {
        suma[i]=suma[i-1]+(str[i]=='a');
        sumb[i]=sumb[i-1]+(str[i]=='b');
    }
    int MAX=0;
    for(int i=1;i<=ans;i++)
        for(int k=i+1;k<=ans;k++)
        {
            MAX=max(MAX,suma[i]/*前面*/+(sumb[k]-sumb[i-1])/*中间*/+(suma[ans]-suma[k-1])/ * Later * / ); 
        } 
    IF (ANS == . 1 ) // Laid arbitration only one case letter 
        COUT << . 1 << endl;
     the else 
        COUT << MAX << endl;
     return  0 ; 
}