#include <bits/stdc++.h>
using namespace std;
bool isPalindrome(string &str, int begin, int end) {
  int mid = (end - begin + 1) / 2;
  for (int i = 0; i < mid; ++i) {
    if (str.at(begin + i) != str.at(end - i)) return false;
  }
  return true;
}
int main() {
  string str;
  cin >> str;
  vector<int> A(str.size(), 0), B(str.size(), 0);
  for (int i = 0; i < str.size(); ++i) {
    for (int j = i; j < str.size(); ++j) {
      if (isPalindrome(str, i, j)) {
        A[j]++;
        B[i]++;
      }
    }
  }
  for (int i = str.size() - 2; i >= 0; --i) {
    B[i] = B[i] + B[i + 1];
  }
  unsigned long long sum = 0;
  for (int i = 0; i < str.size() - 1; ++i) {
    sum += A[i] * B[i + 1];
  }
  cout << sum;
  return 0;
}

### Prompt

Your task is to create a CPP solution to 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;
bool isPalindrome(string &str, int begin, int end) {
  int mid = (end - begin + 1) / 2;
  for (int i = 0; i < mid; ++i) {
    if (str.at(begin + i) != str.at(end - i)) return false;
  }
  return true;
}
int main() {
  string str;
  cin >> str;
  vector<int> A(str.size(), 0), B(str.size(), 0);
  for (int i = 0; i < str.size(); ++i) {
    for (int j = i; j < str.size(); ++j) {
      if (isPalindrome(str, i, j)) {
        A[j]++;
        B[i]++;
      }
    }
  }
  for (int i = str.size() - 2; i >= 0; --i) {
    B[i] = B[i] + B[i + 1];
  }
  unsigned long long sum = 0;
  for (int i = 0; i < str.size() - 1; ++i) {
    sum += A[i] * B[i + 1];
  }
  cout << sum;
  return 0;
}

```