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:
#!/user/bin/python

s = raw_input()
n = len(s)

left = [0 for i in range(n)]
right = [0 for i in range(n)]

for i in range(n) : 
    iterList = [(i, i)]
    if i + 1 != n and \
        s[i] == s[i+1] : 
        iterList.append((i, i+1))

    for l, r in iterList : 
        while l >= 0 and \
            r < n and \
            s[l] == s[r] : 
            left[l] += 1
            right[r] += 1
            l -= 1
            r += 1

cnt = 0
for i in range(n-1) : 
    cnt += right[i] * (sum(left[i+1:]))

print cnt