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;
const double EPS = (1e-10);
bool EQ(double a, double b) { return abs((a) - (b)) < EPS; }
void fast_stream() { std::ios_base::sync_with_stdio(0); }
string str;
bool isPalin[2001][2001];
int finPalin[2001];
int ruiseki[2001];
void solve() {
  cin >> str;
  int n = str.size();
  for (int i = 0; i < n; i++) {
    for (int j = 0;; j++) {
      if (i - j < 0 || i + j >= n) break;
      if (str[i - j] == str[i + j])
        isPalin[i - j][i + j] = true;
      else
        break;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    for (int j = 0;; j++) {
      if (i - j < 0 || i + j + 1 >= n) break;
      if (str[i - j] == str[i + j + 1])
        isPalin[i - j][i + j + 1] = true;
      else
        break;
    }
  }
  for (int i = 0; i < 2001; i++)
    for (int j = 0; j < 2001; j++)
      if (isPalin[i][j]) finPalin[j]++;
  for (int i = n - 1; i >= 0; i--) {
    ruiseki[i] = ruiseki[i + 1];
    for (int j = i; j < n; j++)
      if (isPalin[i][j]) ruiseki[i]++;
  }
  long long res = 0;
  for (int i = 0; i < n - 1; i++) {
    int a = finPalin[i] * ruiseki[i + 1];
    res += a;
  }
  cout << res << endl;
}
int main() {
  solve();
  return 0;
}