#include <bits/stdc++.h>
using namespace std;
long long unsigned A[2000 + 5], B[2000 + 5];
void solve() {
  long long unsigned a, i, j, k, second = 0, sag, sol;
  string str, temp;
  char K[2000 + 5];
  cin >> str;
  for (i = 0; i < str.size(); i++) {
    sag = sol = i;
    while (sol >= 0 && sag < str.size() && str[sol] == str[sag]) {
      A[sol + 1]++;
      B[sag + 1]++;
      sag++;
      sol--;
    }
  }
  for (i = 0; i < str.size(); i++) {
    sag = i + 1;
    sol = i;
    while (sol >= 0 && sag < str.size() && str[sol] == str[sag]) {
      A[sol + 1]++;
      B[sag + 1]++;
      sag++;
      sol--;
    }
  }
  for (i = 1; i <= str.size(); i++) {
    A[i] += A[i - 1];
    B[i] += B[i - 1];
  }
  for (i = 1; i <= str.size(); i++)
    second += (B[i] - B[i - 1]) * (A[str.size()] - A[i]);
  cout << second << endl;
}
int main() {
  solve();
  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;
long long unsigned A[2000 + 5], B[2000 + 5];
void solve() {
  long long unsigned a, i, j, k, second = 0, sag, sol;
  string str, temp;
  char K[2000 + 5];
  cin >> str;
  for (i = 0; i < str.size(); i++) {
    sag = sol = i;
    while (sol >= 0 && sag < str.size() && str[sol] == str[sag]) {
      A[sol + 1]++;
      B[sag + 1]++;
      sag++;
      sol--;
    }
  }
  for (i = 0; i < str.size(); i++) {
    sag = i + 1;
    sol = i;
    while (sol >= 0 && sag < str.size() && str[sol] == str[sag]) {
      A[sol + 1]++;
      B[sag + 1]++;
      sag++;
      sol--;
    }
  }
  for (i = 1; i <= str.size(); i++) {
    A[i] += A[i - 1];
    B[i] += B[i - 1];
  }
  for (i = 1; i <= str.size(); i++)
    second += (B[i] - B[i - 1]) * (A[str.size()] - A[i]);
  cout << second << endl;
}
int main() {
  solve();
  return 0;
}

```