JOIN

 Problem Statement

Problem Statement for PalindromicSubseq2

### Problem Statement

A palindrome is a string that reads the same forwards and backwards. For example, "abba" and "racecar" are palindromes.

An odd palindrome is a palindrome with an odd length. For example, "racecar" is an odd palindrome but "abba" is not. The middle letter of an odd palindrome is called its center.

Limak has a String s consisting of N lowercase English letters.

For each valid i, let X[i] be the number of palidromic subsequences of s such that their center is s[i].

In other words: For a fixed i, there are exactly 2^(N-1) ways to erase some characters of s other than the character s[i]. X[i] is the number of ways of erasing for which the string that remains to the left of s[i] is the reverse of the string that remains to the right of s[i].

For each i, compute the value Y[i] = ((i+1) * X[i]) modulo 1,000,000,007. Then, compute and return the bitwise XOR of all the values Y[i].

### Definition

 Class: PalindromicSubseq2 Method: solve Parameters: String Returns: int Method signature: int solve(String s) (be sure your method is public)

### Constraints

-s will contain exactly N characters.
-N will be between 1 and 3000, inclusive.
-Each character in s will be a lowercase English letter 'a' - 'z'.

### Examples

0)

 `"axbcba"`
`Returns: 31`
 For this string we have X = {1, 2, 2, 4, 2, 1}. Thus, you should return the value (1*1) XOR (2*2) XOR (3*2) XOR (4*4) XOR (5*2) XOR (6*1) = 31. Here is an explanation why X[3] = 4: Given that we want to keep the character s[3], there are 32 possible subsequences of s. Among these there are 4 palindromes centered at s[3]: ```...c.. a..c.a ..bcb. a.bcba ```
1)

 `"eeeee"`
`Returns: 14`
 Here we have X[2] = 6. The six ways of erasing that produce a palindrome centered at s[2] are the following ones: ```..e.. .ee.e .eee. e.e.e e.ee. eeeee ``` Note that we count each way of erasing characters separately, even if different ways of erasing produce the same palindromic string. So, not all X[i] are equal to 6 here.
2)

 `"tcoct"`
`Returns: 4`
 X[] = {1, 2, 4, 2, 1}
3)

 `"zyzyzzzzxzyz"`
`Returns: 221`
4)

 `"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"`
`Returns: 1044407608`
 Note that in this case the answer exceeds the modulo value, because we return the XOR of modulo-ed values, without computing the modulo of that XOR.

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This problem was used for:
Single Round Match 708 Round 1 - Division II, Level Three