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:
from sys import stdin


def arr_sum(arr):
    tem = [0]
    for i in range(len(arr)):
        tem.append(tem[-1] + arr[i])
    return tem


s, ans = stdin.readline().strip(), 0
en, pal = [0] * len(s), [[0] * len(s) for _ in range(len(s))]

for i in range(len(s)):
    for j in range(min(i + 1, len(s) - i)):
        if s[i + j] == s[i - j]:
            en[i + j] += 1
            pal[i - j][i + j] = 1
        else:
            break

for i in range(1, len(s)):
    for j in range(min(i, len(s) - i)):
        if s[i - 1 - j] == s[i + j]:
            en[i + j] += 1
            pal[i - 1 - j][i + j] = 1
        else:
            break

en = arr_sum(en)
for i in range(len(s)):
    for j in range(i, len(s)):
        ans += en[i] * pal[i][j]

print(ans)