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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long Suffix[2020];
vector<long long> v;
int main() {
  string s;
  cin >> s;
  long long n = s.size();
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; i + j < n && i - j >= 0; j++)
      if (s[i + j] == s[i - j])
        v.push_back(i + j), Suffix[i - j]++;
      else
        break;
    for (long long j = 0; i + j + 1 < n && i - j >= 0; j++)
      if (s[i + j + 1] == s[i - j])
        v.push_back(i + j + 1), Suffix[i - j]++;
      else
        break;
  }
  sort(v.begin(), v.end());
  for (long long i = n - 1; i >= 0; i--) Suffix[i] += Suffix[i + 1];
  long long Answer = 0;
  for (long long i = 0; i < v.size(); i++) Answer += Suffix[v[i] + 1];
  cout << Answer;
}