Problem: You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.

In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.

A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.

A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".

Input

The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.

Output

Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.

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

Examples

Input

aa


Output

1


Input

aaa


Output

5


Input

abacaba


Output

36

Solution (in PYTHON3):

Solution:
s = input()
n = len(s)
isPalindrome = [[False for i in range(n)] for j in range(n)]
for i in range(n):
    isPalindrome[i][i] = True
for i in range(n - 1, -1, -1):
    for j in range(i + 1, n):
        if j - i == 1:
            isPalindrome[i][j] = (s[i] == s[j])
        else:
            isPalindrome[i][j] = (s[i] == s[j]) and (isPalindrome[i + 1][j - 1])
pairs = [0] * n
palindromes = [0] * n
palindromes[0] = 1
for i in range(1, n):
    t = 0
    for j in range(1, i + 1):
        if isPalindrome[j][i]:
            pairs[i] += palindromes[j - 1]
            t += 1
    if isPalindrome[0][i]:
        t += 1
    pairs[i] += pairs[i - 1]
    palindromes[i] = palindromes[i - 1] + t
print(pairs[n - 1])