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:
data = raw_input()
n = len(data)
pre = [0]*n
suf = [0]*n

def check(x,y):
    while x<y:
        if data[x]!=data[y]:
            return False
        x+=1
        y-=1
    return True

for i in xrange(0,n):
    l=r=i
    while l>=0 and r<n and data[l] == data[r]:
        pre[l] += 1
        suf[r] += 1
        l-=1
        r+=1
    l=i-1
    r=i
    while l>=0 and r<n and data[l] == data[r]:
        pre[l] += 1
        suf[r] += 1
        l-=1
        r+=1
    '''
    for j in range(i,n):
        if check(i,j):
            pre[i] += 1
            suf[j] += 1
    '''            
for i in xrange(1,n):
    suf[i] += suf[i-1]

#print pre
#print suf
ans = 0
for i in xrange(1,n):
    ans += pre[i]*suf[i-1]

print ans