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;
int main() {
  int a[20001];
  int b[20001];
  for (int i = 0; i < 20001; i++) {
    a[i] = 0;
    b[i] = 0;
  }
  string str;
  cin >> str;
  int gj = str.length();
  for (int i = 1; i < gj; i++) {
    for (int j = 0; j + i < gj; j++) {
      a[j + i]++;
      for (int ii = j, ff = 0; ii < j + i / 2; ii++, ff++)
        if (str[ii] != str[j + i - 1 - ff]) {
          a[j + i]--;
          break;
        }
    }
  }
  string hh = "";
  for (int ii = 0; ii < gj; ii++) hh = str[ii] + hh;
  str = hh;
  for (int i = 1; i < gj; i++) {
    for (int j = 0; j + i < gj; j++) {
      b[j + i]++;
      for (int ii = j, ff = 0; ii < j + i / 2; ii++, ff++)
        if (str[ii] != str[j + i - 1 - ff]) {
          b[j + i]--;
          break;
        }
    }
  }
  long long numb = 0;
  for (int j = 1; j < gj; j++) b[j] = b[j] + b[j - 1];
  for (int i = 1; i < gj; i++) numb = numb + a[i] * b[gj - i];
  cout << numb;
}