CF356E.Xenia and String Problem

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题目描述

Xenia the coder went to The Olympiad of Informatics and got a string problem. Unfortunately, Xenia isn't fabulous in string algorithms. Help her solve the problem.

String ss is a sequence of characters s1s2... sss_{1}s_{2}...\ s_{|s|} , where record s|s| shows the length of the string.

Substring s[i... j]s[i...\ j] of string ss is string sisi+1... sjs_{i}s_{i+1}...\ s_{j} .

String ss is a Gray string, if it meets the conditions:

  • the length of string s|s| is odd;
  • character occurs exactly once in the string;
  • either s=1|s|=1 , or substrings and are the same and are Gray strings.

For example, strings "abacaba", "xzx", "g" are Gray strings and strings "aaa", "xz", "abaxcbc" are not.

The beauty of string pp is the sum of the squares of the lengths of all substrings of string pp that are Gray strings. In other words, consider all pairs of values i,ji,j (1<=i<=j<=p)(1<=i<=j<=|p|) . If substring p[i... j]p[i...\ j] is a Gray string, you should add (ji+1)2(j-i+1)^{2} to the beauty.

Xenia has got string tt consisting of lowercase English letters. She is allowed to replace at most one letter of the string by any other English letter. The task is to get a string of maximum beauty.

输入格式

The first line contains a non-empty string tt (1<=t<=105)(1<=|t|<=10^{5}) . String tt only consists of lowercase English letters.

输出格式

Print the sought maximum beauty value Xenia can get.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

输入输出样例

  • 输入#1

    zzz
    

    输出#1

    12
    
  • 输入#2

    aba
    

    输出#2

    12
    
  • 输入#3

    abacaba
    

    输出#3

    83
    
  • 输入#4

    aaaaaa
    

    输出#4

    15
    

说明/提示

In the first test sample the given string can be transformed into string pp = "zbz". Such string contains Gray strings as substrings p[1... 1]p[1...\ 1] , p[2... 2]p[2...\ 2] , p[3... 3]p[3...\ 3] и p[1... 3]p[1...\ 3] . In total, the beauty of string pp gets equal to 12+12+12+32=121^{2}+1^{2}+1^{2}+3^{2}=12 . You can't obtain a more beautiful string.

In the second test case it is not necessary to perform any operation. The initial string has the maximum possible beauty.

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