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 PYTHON):

Solution:
def main():
    s = raw_input().strip()
    l = len(s) + 1
    pd = []
    for k in xrange(2):
        pd.append([1] * len(s))
    for k in xrange(l):
        ap = [0] * (len(s) - k - 1)
        for i in xrange(1, len(s) - k):
            if s[i-1] == s[i+k] and pd[k][i]: # i, i + k
                ap[i-1] = 1 # i-1, i + k + 1
        pd.append(ap)
    dp_ = [0 for i in xrange(l)]
    for i in xrange(1, l):
        for j in xrange(i):
            dp_[i] += pd[i-j][j] # j, i
    dp = [0 for i in xrange(l)]
    for i in xrange(1, l):
        dp[i] = dp[i-1] + dp_[i]
    for i in xrange(0, len(s)):
        dp_[i] = 0
        for j in xrange(i+1, l):
            dp_[i] += pd[j-i][i] # i, j
    #print dp
    #print dp_
    print sum([dp[i] * dp_[i] for i in xrange(1, len(s))])

main()