Codeforces Round #442 (Div. 2) B - Nikita and string

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.

题意 给出一个只包含a，b 的字符串，可以删除任意数量的字符（不可调换位置），使得ABA的形式的字符串最长（A代表若干个a，B代表若干个b）

求出a，b的前缀和，遍历一遍i，j之间b的个数，找出最大长度

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
#include <set>
using namespace std;

int main()
{
char s[5005];
int a[5005] = {0}, b[5005] = {0};
scanf("%s", s + 1);
int l = strlen(s + 1), A = 0, B = 0, sum = 0;
for (int i = 1; i <= l; i ++) {
b[i] = s[i] == 'b' ? b[i - 1] + 1 : b[i - 1];
a[i] = s[i] == 'a' ? a[i - 1] + 1 : a[i - 1];
}
for (int i = 1; i <= l; i ++) {
for (int j = i - 1; j <= l; j ++) {
if (a[i - 1] + b[j] - b[i - 1] + a[l] - a[j] > sum) {
sum = a[i - 1] + b[j] - b[i - 1] + a[l] - a[j];
}
}
}
printf("%d\n", sum);
}