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:
word = input()
n = len(word)
dp = [[0 for i in range(n)]for j in range(n)]

# 1 size palindrome
for i in range(n):
	dp[i][i] = 1

# i --> size
# j --> start index
for i in range(1,n):
	for j in range(n-i):
		# could be a palindrome
		if word[j]==word[j+i]:
			# size 2 palindrome
			if i-1==0:
				dp[j][j+i] = 1
			else:
				# valid palindrome
				if dp[j+1][j+i-1]==1:
					dp[j][j+i]=1
					
pre = [0 for i in range(n)]
post = [0 for i in range(n)]

# count of palindrome from 1 to i characters
for i in range(n):
	cnt = 0
	for j in range(n):
		cnt+=dp[j][i]
	pre[i] = cnt

# count of palindrome start from ith character
for i in range(n):
	cnt = 0
	for j in range(n):
		cnt+=dp[i][j]
	post[i] = cnt

# prefixing post array 
for i in range(n-2,-1,-1):
	post[i] = post[i]+post[i+1]

# calculating ans 
ans = 0
for i in range(n-1):
	ans+=(pre[i]*post[i+1])
print(ans)