#include <bits/stdc++.h>
using namespace std;
const int max_n = 2010;
int n;
char s[max_n + 10];
int l[max_n + 1] = {0}, r[max_n + 1] = {0}, R[max_n + 1] = {0};
bool P[max_n + 1][max_n + 1];
int main() {
  scanf("%s", s);
  for (int i = 0; true; ++i) {
    if (s[i] == '\0') {
      n = i;
      break;
    } else {
      P[i][i + 1] = true;
      ++l[i + 1];
      ++r[i];
    }
  }
  for (int i = 0; i <= n; ++i) P[i][i] = true;
  for (int k = 2; k <= n; ++k) {
    for (int i = 0; i <= n - k; ++i) {
      int j = i + k;
      P[i][j] = ((P[i + 1][j - 1]) and (s[i] == s[j - 1]));
      if (P[i][j]) {
        ++r[i];
        ++l[j];
      }
    }
  }
  R[n] = 0;
  for (int i = n - 1; i >= 0; --i) R[i] = R[i + 1] + r[i];
  long long res = 0;
  for (int i = 0; i <= n; ++i) res += ((long long)(l[i])) * ((long long)(R[i]));
  cout << res << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
const int max_n = 2010;
int n;
char s[max_n + 10];
int l[max_n + 1] = {0}, r[max_n + 1] = {0}, R[max_n + 1] = {0};
bool P[max_n + 1][max_n + 1];
int main() {
  scanf("%s", s);
  for (int i = 0; true; ++i) {
    if (s[i] == '\0') {
      n = i;
      break;
    } else {
      P[i][i + 1] = true;
      ++l[i + 1];
      ++r[i];
    }
  }
  for (int i = 0; i <= n; ++i) P[i][i] = true;
  for (int k = 2; k <= n; ++k) {
    for (int i = 0; i <= n - k; ++i) {
      int j = i + k;
      P[i][j] = ((P[i + 1][j - 1]) and (s[i] == s[j - 1]));
      if (P[i][j]) {
        ++r[i];
        ++l[j];
      }
    }
  }
  R[n] = 0;
  for (int i = n - 1; i >= 0; --i) R[i] = R[i + 1] + r[i];
  long long res = 0;
  for (int i = 0; i <= n; ++i) res += ((long long)(l[i])) * ((long long)(R[i]));
  cout << res << endl;
  return 0;
}

```