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 answer(n,A):
    p=[[0 for i in range(n)] for j in range(n)]
    
    for i in range(n):
        for j in range(n-i):
            r=j
            c=i+j
            if r==c:
                p[r][c]=1
            
            elif c==r+1:
                if A[c]==A[r]:
                    p[r][c]=1
            else:
                if A[r]==A[c] and p[r+1][c-1]:
                    p[r][c]=1
                    
    s=[0]*n
    s[0]=1
    for i in range(1,n):
        count=0
        for j in range(0,i+1):
            if p[j][i]==1:
                count+=1
        s[i]=s[i-1]+count
        
    dp=[0]*n
    
    for i in range(1,n):
        count=0
        for j in range(i,0,-1):
            count+=p[j][i]*s[j-1]
        
        dp[i]=dp[i-1]+count
        
    
    return dp[-1]
    
s=input()
print(answer(len(s),s))