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

B. Nikita and string

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.

---

---

emmmmmmmm,就是看在不更改各字母顺序的情况下，可以组成形如（a）,（a+b)，(a+b+a)的字符串的最长长度，其实就是子串拼成的a+b+a子串的最长长度，感觉。

ps：可以删除，可以删除，可以删除！……一开始理解有点小问题……一直WA一直WA……orz

这里是dp呀，之前好像有做过类似的？……emmmm不记得了……

拿3个数组记录从s[0]到s[j]的符合条件的最长长度，这三个数组分别表示“a”,"ab"（“b”分到这一情况中）,"aba"（“ba”，(“ab”)分到这一情况中）三种字符串

"a"：只要记录字符串中“a”的个数就好了，每次遇到“a”dp[0][j]++;

"ab"：是由上一种和该情况+b得到的,max(dp[0][j],dp[1][j]),每次遇到"b"++;

"aba"：是由前两种和该情况+a得到的，max(max(dp[0][j],dp[1][j]),dp[2][j]),每次遇到“a”++;

最后输出最长的就好了

  

#include<iostream> 
#include<string>
#include<algorithm>
#include<map>
using namespace std;
string s;
int n,m;
int dp[3][50005];
int main()
{
    while (cin >> s)
    {
        dp[0][0] = 0;
        dp[1][0] = 0;
        dp[2][0] = 0;
        int l = s.length();
        for (int i = 0; i < l; i++)
        {
            dp[0][i + 1] = dp[0][i] + (s[i] == 'a');
            dp[1][i + 1] = max(dp[0][i], dp[1][i]) + (s[i] == 'b');
            dp[2][i + 1] = max(max(dp[0][i],dp[1][i]), dp[2][i]) + (s[i] == 'a');
        }
        cout << max(max(dp[0][l], dp[1][l]), dp[2][l]) << endl;
    }
    return 0;
}