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 PYTHON3):

Solution:
def ispalindrome(s):
	if s==s[::-1]:
		return True
	return False

s = input()
n = len(s)
dp = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
	dp[i][i] = 1
for i in range(1,n):
	for j in range(i-1,-1,-1):
		if s[i]==s[j] and (j+1>i-1 or dp[j+1][i-1]):
			dp[j][i] = 1
pref = [0]*n
suff = [0]*n
pref[0] = suff[-1] = 1
for i in range(1,n):
	# pref[i] = pref[i-1]
	for j in range(i,-1,-1):
		if dp[j][i]:
			pref[i] += 1
for i in range(n-2,-1,-1):
	suff[i] = suff[i+1]
	for j in range(i,n):
		if dp[i][j]:
			suff[i] += 1

# for i in dp:
# 	print (*i)
# print (pref)
# print (suff)

ans = 0
curr = 0
for i in range(n-1):
	ans += suff[i+1]*pref[i]
print (ans)