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:
s = input()
# s = "aiajadabaa"
n = len(s)

L = [0] * n  # L[i] = số lượng xâu đối xứng trong đoạn s[0..i]
R = [0] * n  # R[i] = số lượng xâu đối xứng trong đoạn s[i+1..n-1]

for i in range(n):
	x = y = i
	while x >= 0 and y < n and s[x] == s[y]:
		R[x] += 1
		L[y] += 1
		x -= 1
		y += 1
for i in range(1, n):
	if s[i - 1] == s[i]:
		x, y = i-1, i
		while x >= 0 and y < n and s[x] == s[y]:
			R[x] += 1
			L[y] += 1
			x -= 1
			y += 1

for i in range(n - 1):
	L[i + 1] += L[i]
	
p = 0
for i in range(1, n):
	p += L[i-1] * R[i]

print(p)