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 in;
bool dpp[2005][2005];
unsigned long long dp2[2005][2005] = {0};
bool ispal(int s, int e) {
  for (; s < e; s++, e--)
    if (in[s] != in[e]) return false;
  return true;
}
int main() {
  cin >> in;
  unsigned long long w = 0;
  for (int ps = 1; ps <= in.length(); ps++)
    for (int i = 0; i <= in.length(); i++)
      dpp[i][ps] = (i + ps > in.length()) ? 0 : ispal(i, i + ps - 1);
  for (int ps = 1; ps <= in.length(); ps++)
    for (int i = in.length(); i >= 0; i--)
      dp2[i][ps] += dp2[i + 1][ps] + dpp[i][ps];
  for (int ps = 1; ps <= in.length(); ps++)
    for (int i = 0; i <= in.length(); i++) dp2[i][ps] += dp2[i][ps - 1];
  for (int ps = 1; ps < in.length(); ps++)
    for (int i = 0; i < in.length() - ps; i++)
      if (dpp[i][ps]) w += dp2[i + ps][in.length()];
  cout << w << endl;
  return 0;
}