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 s is a sequence of characters s1s2... s|s|, where record |s| shows the length of the string. Substring s[i... j] of string s is string sisi+1... sj.String s is a Gray string, if it meets the conditions: the length of string |s| is odd; character occurs exactly once in the string; either |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 p is the sum of the squares of the lengths of all substrings of string p that are Gray strings. In other words, consider all pairs of values i,j (1 \le i \le j \le |p|). If substring p[i... j] is a Gray string, you should add (j-i+1)2 to the beauty.Xenia has got string t 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.

Input The first line contains a non-empty string t (1 \le |t| \le 105). String t only consists of lowercase English letters.

Output Print the sought maximum beauty value Xenia can get.Please do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

In the first test sample the given string can be transformed into string p = "zbz". Such string contains Gray strings as substrings p[1... 1], p[2... 2], p[3... 3] , p[1... 3]. In total, the beauty of string p gets equal to 12+12+12+32=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.