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 = list(raw_input())
N = len(s)
pre = [1 for x in range(N)]
suf = [1 for x in range(N)]
for i in range(N):
	for j in range(min(i,N-i-1)):
		if s[i-j-1] == s[i+j+1]:
			pre[i-j-1] += 1
			suf[i+j+1] += 1
		else:
			break
	if i and s[i]==s[i-1]:
		pre[i-1] += 1
		suf[i] += 1
		for j in range(min(i-1,N-i-1)):
			if s[i-j-2] == s[i+j+1]:
				pre[i-j-2] += 1
				suf[i+j+1] += 1
			else:
				break
cur = 0
ans = 0
for i in range(N):
	ans += cur * pre[i]
	cur += suf[i]
print ans