#include <bits/stdc++.h>
using namespace std;
int arr[2005][2005];
int brr[2005];
int crr[2005];
string s;
long long f(int n) {
  for (int len = 1; len <= n; len++) {
    for (int i = 1; i <= n - len + 1; i++) {
      int j = i + len - 1;
      if (i == j)
        arr[i][j] = 1;
      else if (j == i + 1) {
        if (s[i] == s[j])
          arr[i][j] = 1;
        else
          arr[i][j] = 0;
      } else {
        if ((s[i] == s[j]) && arr[i + 1][j - 1] == 1)
          arr[i][j] = 1;
        else
          arr[i][j] = 0;
      }
    }
  }
  for (int i = n; i >= 1; i--) {
    brr[i] = 0;
    for (int j = i; j <= n; j++) {
      if (arr[i][j]) brr[i]++;
    }
  }
  long long sum = 0;
  long long cummulative = 0;
  for (int i = 1; i <= n; i++) {
    sum += 1LL * cummulative * brr[i];
    for (int j = 1; j <= i; j++) {
      if (arr[j][i]) cummulative++;
    }
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> s;
  int siz = s.size();
  s = " " + s;
  cout << f(siz) << '\n';
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 arr[2005][2005];
int brr[2005];
int crr[2005];
string s;
long long f(int n) {
  for (int len = 1; len <= n; len++) {
    for (int i = 1; i <= n - len + 1; i++) {
      int j = i + len - 1;
      if (i == j)
        arr[i][j] = 1;
      else if (j == i + 1) {
        if (s[i] == s[j])
          arr[i][j] = 1;
        else
          arr[i][j] = 0;
      } else {
        if ((s[i] == s[j]) && arr[i + 1][j - 1] == 1)
          arr[i][j] = 1;
        else
          arr[i][j] = 0;
      }
    }
  }
  for (int i = n; i >= 1; i--) {
    brr[i] = 0;
    for (int j = i; j <= n; j++) {
      if (arr[i][j]) brr[i]++;
    }
  }
  long long sum = 0;
  long long cummulative = 0;
  for (int i = 1; i <= n; i++) {
    sum += 1LL * cummulative * brr[i];
    for (int j = 1; j <= i; j++) {
      if (arr[j][i]) cummulative++;
    }
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> s;
  int siz = s.size();
  s = " " + s;
  cout << f(siz) << '\n';
}

```