#include <bits/stdc++.h>
using namespace std;
int main() {
  string s;
  cin >> s;
  int n = s.size();
  s = "~" + s + "/";
  vector<int> v[2012];
  for (int i = 1; i <= n; i++) {
    int l = i, r = i;
    while (s[l] == s[r]) v[l].push_back(r), --l, ++r;
  }
  for (int i = 1; i <= n - 1; i++) {
    int l = i, r = i + 1;
    while (s[l] == s[r]) v[l].push_back(r), --l, ++r;
  }
  long long homos[2012] = {};
  long long ans = 0, sum = 0;
  for (int i = 1; i <= n; i++) {
    sum += homos[i];
    ans += v[i].size() * sum;
    for (int j = 0; j < v[i].size(); j++) homos[v[i][j] + 1] += 1;
  }
  cout << ans << endl;
}

### Prompt

Generate 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;
int main() {
  string s;
  cin >> s;
  int n = s.size();
  s = "~" + s + "/";
  vector<int> v[2012];
  for (int i = 1; i <= n; i++) {
    int l = i, r = i;
    while (s[l] == s[r]) v[l].push_back(r), --l, ++r;
  }
  for (int i = 1; i <= n - 1; i++) {
    int l = i, r = i + 1;
    while (s[l] == s[r]) v[l].push_back(r), --l, ++r;
  }
  long long homos[2012] = {};
  long long ans = 0, sum = 0;
  for (int i = 1; i <= n; i++) {
    sum += homos[i];
    ans += v[i].size() * sum;
    for (int j = 0; j < v[i].size(); j++) homos[v[i][j] + 1] += 1;
  }
  cout << ans << endl;
}

```