#include <bits/stdc++.h>
using namespace std;
bool mat[2001][2001];
string s;
void mark(int l, int r) {
  int sz = s.size();
  while (l >= 0 and r < sz and s[r] == s[l]) {
    mat[l][r] = 1;
    l--;
    r++;
  }
}
vector<int> x, y;
int main() {
  cin >> s;
  int sz = s.size();
  memset(mat, 0, sizeof mat);
  for (int i = 0; i < sz; i++) {
    mark(i, i);
    mark(i - 1, i);
    mark(i, i + 1);
  }
  for (int i = 0; i < sz; i++)
    for (int j = 0; j < sz; j++)
      if (mat[i][j]) x.push_back(i), y.push_back(j);
  long long ans = 0;
  for (int i = 0; i < x.size(); i++) {
    int u = upper_bound(x.begin(), x.end(), y[i]) - x.begin();
    ans += x.size() - u;
  }
  cout << ans << endl;
}

### 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;
bool mat[2001][2001];
string s;
void mark(int l, int r) {
  int sz = s.size();
  while (l >= 0 and r < sz and s[r] == s[l]) {
    mat[l][r] = 1;
    l--;
    r++;
  }
}
vector<int> x, y;
int main() {
  cin >> s;
  int sz = s.size();
  memset(mat, 0, sizeof mat);
  for (int i = 0; i < sz; i++) {
    mark(i, i);
    mark(i - 1, i);
    mark(i, i + 1);
  }
  for (int i = 0; i < sz; i++)
    for (int j = 0; j < sz; j++)
      if (mat[i][j]) x.push_back(i), y.push_back(j);
  long long ans = 0;
  for (int i = 0; i < x.size(); i++) {
    int u = upper_bound(x.begin(), x.end(), y[i]) - x.begin();
    ans += x.size() - u;
  }
  cout << ans << endl;
}

```