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:
from collections import defaultdict
S = input()
N = len(S)
# Step 1: Find the positions of a-z
Pos = defaultdict(list)
for i in range(N):
    Pos[S[i]].append(i)
# Step 2: Find Ends here dp
# Minor Change: Add a bool dict to catch the palindrome
Range = defaultdict(bool)
dpE = [1] * N
for i in range(N):
    Range[(i,i)] = True
    for pos in Pos[S[i]]:
        if(pos>=i):
            break
        else:
            if(i - pos == 1):
                if(S[i] == S[pos]):
                    Range[(pos,i)] = True
                    dpE[i] += 1
            elif(Range[(pos+1,i-1)]):
                Range[(pos,i)] = True
                dpE[i] += 1

# Step 3: Find  Till Here dp

dpT = [0] * N
dpT[-1] = 1
for i in range(N-2,-1,-1):
    dpT[i] += dpT[i+1] + 1
    for j in range(N-1,i,-1):
        if(Range[(i,j)]):
            dpT[i] += 1

# Step 4: It's Time
Ans = 0
for i in range(N-1):
    Ans += dpE[i] * dpT[i+1]
print(Ans)