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:
import sys
import copy



def min(a, b):
    if a < b:
        return a
    return b


def count_left(n, d1, d2):
    sum = 0
    for i in xrange(n+1):
        if d1[i] >=  n - i + 1:
            sum += 1
        if d2[i] >= n - i + 1:
            sum += 1
    return sum

def count_right(n, d1, d2):
    sum = 0
    for i in xrange(n+1, len(d1)):
        sum += min(d1[i], i - n)
        try:
            sum += min(d2[i + 1], i - n)
        except:
            pass
    return sum



def podpal_array(s):
    d1 = []
    n = len(s)
    for i in xrange(n):
        d1.append(0)

    l = 0
    r = -1
    for i in xrange(n):
        if i>r:
            k = 1
        else:
            k = min (copy.copy(d1[l+r-i]), r-i) + 1
        while i + k < n and i - k >= 0 and s[i+k] == s[i-k]:
            k += 1
        k = k - 1
        d1[i] = k
        if (i + k > r):
            l = i - k
            r = i + k
    for x in xrange(len(d1)):
        d1[x] += 1

    d2 = []
    n = len(s)
    for i in xrange(n):
        d2.append(0)

    l = 0
    r = -1
    for i in xrange(n):
        if i>r:
            k = 1
        else:
            k = min (copy.copy(d2[l+r-i + 1]), r-i + 1) + 1
        while i + k - 1 < n and i - k >= 0 and s[i + k - 1] == s[i - k]:
            k += 1
        k = k - 1
        d2[i] = k
        if (i + k > r):
            l = i - k
            r = i + k - 1
    
    return d1, d2

def main():
    s = sys.stdin.readline().split()[0]
    d1, d2 = podpal_array(s)
    
    sum = 0
    for i in xrange(len(s)-1):
        s1 = count_left(i, d1, d2)
        s2 = count_right(i, d1, d2)
        sum += s1 * s2
    print sum

if __name__ == "__main__":
    main()