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 as I

s = I.readline()[:-1]
n = len(s)
st = [1]*n
en = [1]*n

# Odd length
for i in xrange(n):
    d = 1
    while 0 <= i-d and i+d < n:
        if s[i-d] == s[i+d]:
            st[i-d] += 1
            en[i+d] += 1
            d += 1
        else:
            break
            
# Even length
for i in xrange(n-1):
    if s[i] == s[i+1]:
        st[i] += 1
        en[i+1] += 1
        d = 1
        while 0 <= i-d and i+d+1 < n:
            if s[i-d] == s[i+d+1]:
                st[i-d] += 1
                en[i+d+1] += 1
                d += 1
            else:
                break

# Partial sum                
for i in xrange(1, n):
    #en[i] += en[i-1]
    st[n-i-1] += st[n-i]
    
# Totals
rv = 0
for i in xrange(n-1):
    rv += en[i] * st[i+1]

print rv