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()
n = len(s)

begin = []
end = []

for i in xrange(n):
    begin.append(0)
    end.append(0)


for i in xrange(n):
    begin[i] = begin[i] + 1
    end[i] = end[i] + 1
    l = i - 1
    r = i + 1
    while( l >= 0 and r < n and s[l] == s[r]):
        begin[l] = begin[l] + 1
        end[r] = end[r] + 1
        l-=1
        r+=1
    l = i
    r = i + 1
    while( l >=0 and r < n and s[l] == s[r]):
        begin[l] = begin[l] + 1
        end[r] = end[r] + 1        
        l-=1
        r+=1

ans = 0

for i in xrange(n):
    for j in xrange(i+1,n):
        ans = ans + end[i] * begin[j];

print ans