#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
string t;
void solve() {
  int n = t.size();
  vector<int> d(n, 0), s(n, 0);
  for (int i = 0; i < n; i++) {
    int j = i, k = i;
    while (0 <= j && k < n && t[j] == t[k]) {
      s[j]++;
      d[k]++;
      j--;
      k++;
    }
    j = i - 1;
    k = i;
    while (0 <= j && k < n && t[j] == t[k]) {
      s[j]++;
      d[k]++;
      j--;
      k++;
    }
  }
  vector<long long> p(n, 0);
  p[0] = d[0];
  for (int i = 1; i < n; i++) {
    p[i] = p[i - 1] + d[i];
  }
  long long r = 0;
  for (int i = 1; i < n; i++) {
    r += p[i - 1] * s[i];
  }
  cout << r << endl;
}
int main() {
  while (cin >> t) {
    solve();
  }
}

### Prompt

Your task is to create 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 double pi = acos(-1.0);
string t;
void solve() {
  int n = t.size();
  vector<int> d(n, 0), s(n, 0);
  for (int i = 0; i < n; i++) {
    int j = i, k = i;
    while (0 <= j && k < n && t[j] == t[k]) {
      s[j]++;
      d[k]++;
      j--;
      k++;
    }
    j = i - 1;
    k = i;
    while (0 <= j && k < n && t[j] == t[k]) {
      s[j]++;
      d[k]++;
      j--;
      k++;
    }
  }
  vector<long long> p(n, 0);
  p[0] = d[0];
  for (int i = 1; i < n; i++) {
    p[i] = p[i - 1] + d[i];
  }
  long long r = 0;
  for (int i = 1; i < n; i++) {
    r += p[i - 1] * s[i];
  }
  cout << r << endl;
}
int main() {
  while (cin >> t) {
    solve();
  }
}

```