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:
t = input()
n = len(t)
k = n // 2
a, b = [1] * n, [1] * n
for x, y in [(i, i + 2) for i in range(k - 1)]:
    while x >= 0 and t[x] == t[y]:
        a[y] += 1
        b[x] += 1
        x -= 1
        y += 1
for x, y in [(i, i + 1) for i in range(k)]:
    while x >= 0 and t[x] == t[y]:
        a[y] += 1
        b[x] += 1
        x -= 1
        y += 1
for x, y in [(i, i + 2) for i in range(k - 1, n - 1)]:
    while y < n and t[x] == t[y]:
        a[y] += 1
        b[x] += 1
        x -= 1
        y += 1
for x, y in [(i, i + 1) for i in range(k, n)]:
    while y < n and t[x] == t[y]:
        a[y] += 1
        b[x] += 1
        x -= 1
        y += 1
for i in range(n - 1):
    a[i + 1] += a[i]
print(sum(a[i] * b[i + 1] for i in range(n - 1)))