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:
s=raw_input()
if s.count(s[0]) == len(s):
    n=len(s)-1
    print  (6*n+11*n*n+6*n**3+n**4)/24
    exit()
d=[0]*len(s)
r=[0]*len(s)
for i in range(len(s)):
    for j in [0,1]:
        for k in range(len(s)):
            if i-k >= 0 and i+k+j < len(s) and s[i-k] == s[i+k+j]:
                d[i+k+j] += 1
                r[i-k] += 1                        
            else:
                break     

s=0
c=0
for i, e in reversed(list(enumerate(r))[1:]):
   
    s += e
    #print s*d[i-1]
    c += s*d[i-1]
    
print c