#include <bits/stdc++.h>
using namespace std;
string s;
bool isPal(long long int i, long long int j) {
  for (long long int k = i; k < (i + j + 1) / 2; k++)
    if (s[k] != s[j + i - k]) return 0;
  return 1;
}
bool is[2010][2010];
long long int nd[2010], sm[2010];
int main() {
  cin >> s;
  long long int sz = s.size();
  for (long long int i = 0; i < sz; i++)
    for (long long int j = i; j < sz; j++) {
      is[i][j] = isPal(i, j);
      if (is[i][j]) nd[j]++;
    }
  sm[0] = nd[0];
  for (long long int i = 1; i < sz; i++) sm[i] = sm[i - 1] + nd[i];
  long long int ans = 0;
  for (long long int i = 1; i < sz; i++) {
    for (long long int j = i; j < sz; j++) {
      ans += is[i][j] * sm[i - 1];
    }
  }
  cout << ans << endl;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
string s;
bool isPal(long long int i, long long int j) {
  for (long long int k = i; k < (i + j + 1) / 2; k++)
    if (s[k] != s[j + i - k]) return 0;
  return 1;
}
bool is[2010][2010];
long long int nd[2010], sm[2010];
int main() {
  cin >> s;
  long long int sz = s.size();
  for (long long int i = 0; i < sz; i++)
    for (long long int j = i; j < sz; j++) {
      is[i][j] = isPal(i, j);
      if (is[i][j]) nd[j]++;
    }
  sm[0] = nd[0];
  for (long long int i = 1; i < sz; i++) sm[i] = sm[i - 1] + nd[i];
  long long int ans = 0;
  for (long long int i = 1; i < sz; i++) {
    for (long long int j = i; j < sz; j++) {
      ans += is[i][j] * sm[i - 1];
    }
  }
  cout << ans << endl;
}

```