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;
string s;
long long _left[2000];
long long _right[2000];
bool is_pol(int a, int b) {
  int l = (b - a) / 2 + (b - a) % 2;
  for (int i = 0; i < l; i++)
    if (s[a + i] != s[b - i]) return false;
  return true;
}
int number_of_left_pols(int x) {
  int result = 1;
  for (int i = 0; i < x; i++)
    if (is_pol(i, x)) result++;
  return result;
}
int number_of_right_pols(int x) {
  int result = 1;
  for (int i = x + 1; i < s.size(); i++)
    if (is_pol(x, i)) result++;
  return result;
}
int main() {
  long long ans;
  cin >> s;
  if (s.size() == 1) {
    cout << 0;
    return 0;
  } else if (s.size() == 2) {
    cout << 1;
    return 0;
  }
  _left[0] = 1;
  for (int i = 1; i < s.size(); i++)
    _left[i] = _left[i - 1] + number_of_left_pols(i);
  _right[s.size() - 1] = 1;
  for (int i = s.size() - 2; i >= 0; i--)
    _right[i] = _right[i + 1] + number_of_right_pols(i);
  ans = _right[1];
  for (int i = 1; i < s.size() - 1; i++)
    ans += (_left[i] - _left[i - 1]) * _right[i + 1];
  cout << ans;
  return 0;
}