#include <bits/stdc++.h>
using namespace std;
map<int, int> T;
long long res = 0;
string S;
long long R[2048];
int main() {
  cin >> S;
  int N = S.size();
  memset(R, 0, sizeof R);
  for (int offset = 0; offset < (2); offset++) {
    for (int i = 0; i < (N); i++) {
      int left = i, right = i + offset;
      while (left >= 0 && right < N) {
        if (S[left] == S[right]) {
          R[left]++;
          T[right]++;
        } else {
          break;
        }
        left--;
        right++;
      }
    }
  }
  for (int i = 2010; i >= 0; i--) {
    R[i] += R[i + 1];
  }
  for (int i = 0; i < (2040); i++) {
    res += R[i + 1] * T[i];
  }
  cout << res << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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;
map<int, int> T;
long long res = 0;
string S;
long long R[2048];
int main() {
  cin >> S;
  int N = S.size();
  memset(R, 0, sizeof R);
  for (int offset = 0; offset < (2); offset++) {
    for (int i = 0; i < (N); i++) {
      int left = i, right = i + offset;
      while (left >= 0 && right < N) {
        if (S[left] == S[right]) {
          R[left]++;
          T[right]++;
        } else {
          break;
        }
        left--;
        right++;
      }
    }
  }
  for (int i = 2010; i >= 0; i--) {
    R[i] += R[i + 1];
  }
  for (int i = 0; i < (2040); i++) {
    res += R[i + 1] * T[i];
  }
  cout << res << endl;
  return 0;
}

```