#include <bits/stdc++.h>
using namespace std;
int main() {
  string s;
  cin >> s;
  int siz = s.size(), i = 0;
  int as[2010], de[2010];
  memset(as, 0, sizeof(as));
  memset(de, 0, sizeof(de));
  while (i < siz) {
    int j = i, k = i;
    while (j < siz && k >= 0) {
      if (s[j] == s[k]) {
        as[k]++;
        de[j]++;
      } else
        break;
      j++;
      k--;
    }
    j = i + 1, k = i;
    while (j < siz && k >= 0) {
      if (s[j] == s[k]) {
        as[k]++;
        de[j]++;
      } else
        break;
      j++;
      k--;
    }
    i++;
  }
  i = 0;
  long long sum = 0;
  while (i < siz) {
    int j = i + 1;
    while (j < siz) {
      sum += (de[i] * as[j]);
      j++;
    }
    i++;
  }
  cout << sum << 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 main() {
  string s;
  cin >> s;
  int siz = s.size(), i = 0;
  int as[2010], de[2010];
  memset(as, 0, sizeof(as));
  memset(de, 0, sizeof(de));
  while (i < siz) {
    int j = i, k = i;
    while (j < siz && k >= 0) {
      if (s[j] == s[k]) {
        as[k]++;
        de[j]++;
      } else
        break;
      j++;
      k--;
    }
    j = i + 1, k = i;
    while (j < siz && k >= 0) {
      if (s[j] == s[k]) {
        as[k]++;
        de[j]++;
      } else
        break;
      j++;
      k--;
    }
    i++;
  }
  i = 0;
  long long sum = 0;
  while (i < siz) {
    int j = i + 1;
    while (j < siz) {
      sum += (de[i] * as[j]);
      j++;
    }
    i++;
  }
  cout << sum << endl;
  return 0;
}

```