#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > v;
int d[2002];
int f[2002];
string s;
bool IsPal(int a, int b) {
  for (int i = 0; i < (b - a + 1) / 2; i++) {
    if (s[a + i] != s[b - i]) return false;
  }
  return true;
}
int main() {
  cin >> s;
  int n = s.size();
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (IsPal(i, j)) {
        v.push_back({i, j});
      }
    }
  }
  for (int i = 0; i < v.size(); i++) {
    d[v[i].first]++;
    f[v[i].second]++;
  }
  long long sol = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      sol += f[i] * d[j];
    }
  }
  cout << sol;
}

### Prompt

Please formulate 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;
vector<pair<int, int> > v;
int d[2002];
int f[2002];
string s;
bool IsPal(int a, int b) {
  for (int i = 0; i < (b - a + 1) / 2; i++) {
    if (s[a + i] != s[b - i]) return false;
  }
  return true;
}
int main() {
  cin >> s;
  int n = s.size();
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (IsPal(i, j)) {
        v.push_back({i, j});
      }
    }
  }
  for (int i = 0; i < v.size(); i++) {
    d[v[i].first]++;
    f[v[i].second]++;
  }
  long long sol = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      sol += f[i] * d[j];
    }
  }
  cout << sol;
}

```