#include <bits/stdc++.h>
using namespace std;
int main() {
  string second;
  cin >> second;
  vector<int> st, en;
  int p = 0, q = 0;
  for (int cnt = 0;; cnt++) {
    int i = p;
    int j = q;
    if (cnt % 2 == 0)
      q++;
    else
      p++;
    while (i >= 0 && j < second.size()) {
      if (second[i] != second[j]) break;
      st.push_back(i);
      en.push_back(j);
      i--;
      j++;
    }
    if (q == second.size()) break;
  }
  sort(st.begin(), st.end());
  sort(en.begin(), en.end());
  long long ans = 0;
  for (int i = 0; i < en.size(); i++) {
    long long it = upper_bound(st.begin(), st.end(), en[i]) - st.begin();
    ans += (long long)st.size() - it;
  }
  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 second;
  cin >> second;
  vector<int> st, en;
  int p = 0, q = 0;
  for (int cnt = 0;; cnt++) {
    int i = p;
    int j = q;
    if (cnt % 2 == 0)
      q++;
    else
      p++;
    while (i >= 0 && j < second.size()) {
      if (second[i] != second[j]) break;
      st.push_back(i);
      en.push_back(j);
      i--;
      j++;
    }
    if (q == second.size()) break;
  }
  sort(st.begin(), st.end());
  sort(en.begin(), en.end());
  long long ans = 0;
  for (int i = 0; i < en.size(); i++) {
    long long it = upper_bound(st.begin(), st.end(), en[i]) - st.begin();
    ans += (long long)st.size() - it;
  }
  cout << ans << endl;
}

```