#include <bits/stdc++.h>
using namespace std;
int dp[2][2010];
int main() {
  string a;
  cin >> a;
  int l = a.size();
  for (int i = 0; i < l; i++) {
    int j = 0;
    while (i - j >= 0 && i + j < l)
      if (a[i - j] == a[i + j]) {
        dp[0][i - j]++;
        dp[1][i + j]++;
        j++;
      } else
        break;
  }
  for (int i = 0; i < l - 1; i++) {
    int j = 0;
    while (i - j >= 0 && i + j + 1 < l)
      if (a[i - j] == a[i + j + 1]) {
        dp[0][i - j]++;
        dp[1][i + j + 1]++;
        j++;
      } else
        break;
  }
  for (int i = 0; i < l; i++) dp[0][l - i - 1] += dp[0][l - i];
  long long cnt = 0;
  for (int i = 0; i < l - 1; i++) cnt += (dp[0][i + 1] * dp[1][i]);
  cout << cnt << endl;
  return 0;
}

### Prompt

Generate 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;
int dp[2][2010];
int main() {
  string a;
  cin >> a;
  int l = a.size();
  for (int i = 0; i < l; i++) {
    int j = 0;
    while (i - j >= 0 && i + j < l)
      if (a[i - j] == a[i + j]) {
        dp[0][i - j]++;
        dp[1][i + j]++;
        j++;
      } else
        break;
  }
  for (int i = 0; i < l - 1; i++) {
    int j = 0;
    while (i - j >= 0 && i + j + 1 < l)
      if (a[i - j] == a[i + j + 1]) {
        dp[0][i - j]++;
        dp[1][i + j + 1]++;
        j++;
      } else
        break;
  }
  for (int i = 0; i < l; i++) dp[0][l - i - 1] += dp[0][l - i];
  long long cnt = 0;
  for (int i = 0; i < l - 1; i++) cnt += (dp[0][i + 1] * dp[1][i]);
  cout << cnt << endl;
  return 0;
}

```