#include <bits/stdc++.h>
using namespace std;
int countBits(long long mask) {
  int res = 0;
  while (mask) mask &= (mask - 1), ++res;
  return res;
}
string toString(long long n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
long long toNumber(string s) {
  stringstream ss;
  long long n;
  ss << s;
  ss >> n;
  return n;
}
bool palindrome[2001][2001];
int Start[2001], End[2001], cumEnd[2001];
int main() {
  string str;
  while (cin >> str) {
    memset(Start, 0, sizeof Start);
    memset(End, 0, sizeof End);
    memset(palindrome, 0, sizeof palindrome);
    for (int i = (int)(0), _m = (int)(str.size()); i < _m; ++i) {
      palindrome[i][i] = true;
      if (i && str[i] == str[i - 1]) palindrome[i - 1][i] = true;
    }
    for (int len = (int)(3), _m = (int)(str.size()); len < _m; ++len)
      for (int i = (int)(0), _m = (int)(str.size()); i < _m; ++i)
        if (len + i - 1 >= str.size())
          break;
        else
          palindrome[i][i + len - 1] =
              (str[i] == str[i + len - 1]) && palindrome[i + 1][i + len - 2];
    for (int i = (int)(0), _m = (int)(str.size()); i < _m; ++i) {
      for (int j = (int)(i), _m = (int)(str.size()); j < _m; ++j)
        Start[i] += palindrome[i][j];
      for (int j = (int)(0), _m = (int)(i + 1); j < _m; ++j)
        End[i] += palindrome[j][i];
      cumEnd[i] = End[i] + (!i ? 0 : cumEnd[i - 1]);
    }
    long long res = 0;
    for (int i = (int)(1), _m = (int)(str.size()); i < _m; ++i)
      res += (long long)Start[i] * cumEnd[i - 1];
    cout << res << endl;
  }
}

### Prompt

In Cpp, your task is to solve the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int countBits(long long mask) {
  int res = 0;
  while (mask) mask &= (mask - 1), ++res;
  return res;
}
string toString(long long n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
long long toNumber(string s) {
  stringstream ss;
  long long n;
  ss << s;
  ss >> n;
  return n;
}
bool palindrome[2001][2001];
int Start[2001], End[2001], cumEnd[2001];
int main() {
  string str;
  while (cin >> str) {
    memset(Start, 0, sizeof Start);
    memset(End, 0, sizeof End);
    memset(palindrome, 0, sizeof palindrome);
    for (int i = (int)(0), _m = (int)(str.size()); i < _m; ++i) {
      palindrome[i][i] = true;
      if (i && str[i] == str[i - 1]) palindrome[i - 1][i] = true;
    }
    for (int len = (int)(3), _m = (int)(str.size()); len < _m; ++len)
      for (int i = (int)(0), _m = (int)(str.size()); i < _m; ++i)
        if (len + i - 1 >= str.size())
          break;
        else
          palindrome[i][i + len - 1] =
              (str[i] == str[i + len - 1]) && palindrome[i + 1][i + len - 2];
    for (int i = (int)(0), _m = (int)(str.size()); i < _m; ++i) {
      for (int j = (int)(i), _m = (int)(str.size()); j < _m; ++j)
        Start[i] += palindrome[i][j];
      for (int j = (int)(0), _m = (int)(i + 1); j < _m; ++j)
        End[i] += palindrome[j][i];
      cumEnd[i] = End[i] + (!i ? 0 : cumEnd[i - 1]);
    }
    long long res = 0;
    for (int i = (int)(1), _m = (int)(str.size()); i < _m; ++i)
      res += (long long)Start[i] * cumEnd[i - 1];
    cout << res << endl;
  }
}

```