#include <bits/stdc++.h>
using namespace std;
void print(vector<int> &v) {
  for (int i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << endl;
}
string s;
bool check(vector<vector<int> > &v, int from, int to) {
  while (from < to) {
    if (s[from] != s[to]) {
      return false;
    }
    from++;
    to--;
  }
  return 1;
}
int sum1(vector<vector<int> > &v, int i, int j) {
  int cnt = 0;
  for (j; j >= 0; j--) {
    if (v[j][i]) cnt++;
  }
  return cnt;
}
int sum2(vector<vector<int> > &v, int i, int j) {
  int cnt = 0;
  for (j; j < v.size(); j++) {
    if (v[i][j]) cnt++;
  }
  return cnt;
}
void solve() {
  cin >> s;
  map<char, set<int> > mp;
  vector<vector<int> > sm;
  vector<int> tmp(26, 0);
  sm.push_back(tmp);
  for (int i = 0; i < s.length(); i++) {
    mp[s[i]].insert(i);
    sm.push_back(sm[sm.size() - 1]);
    sm[sm.size() - 1][s[i] - 'a']++;
  }
  tmp.assign(s.size(), 0);
  vector<vector<int> > palindroms(s.length(), tmp);
  for (int i = 0; i < s.length(); i++) {
    for (int j = i; j < s.length(); j++) {
      if (check(sm, i, j)) palindroms[i][j] = 1;
    }
  }
  long long int oldLeft = 0;
  long long int oldRight = 0;
  for (int i = 0; i < s.length(); i++) {
    for (int j = i; j < s.length(); j++) oldRight += palindroms[i][j];
  }
  long long int answ = 0;
  for (int i = 0; i < s.length() - 1; i++) {
    int d1 = 0, d2 = 0;
    for (int j = 0; j <= i; j++) {
      if (palindroms[j][i]) d1++;
    }
    for (int j = i; j < s.length(); j++) {
      if (palindroms[i][j]) d2++;
    }
    answ += d1 * (oldRight - d2);
    oldRight -= d2;
    oldLeft += d1;
  }
  cout << answ << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 0;
  if (t == 1)
    cin >> t;
  else
    t = 1;
  while (t--) solve();
}

### Prompt

Create a solution in cpp for 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;
void print(vector<int> &v) {
  for (int i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << endl;
}
string s;
bool check(vector<vector<int> > &v, int from, int to) {
  while (from < to) {
    if (s[from] != s[to]) {
      return false;
    }
    from++;
    to--;
  }
  return 1;
}
int sum1(vector<vector<int> > &v, int i, int j) {
  int cnt = 0;
  for (j; j >= 0; j--) {
    if (v[j][i]) cnt++;
  }
  return cnt;
}
int sum2(vector<vector<int> > &v, int i, int j) {
  int cnt = 0;
  for (j; j < v.size(); j++) {
    if (v[i][j]) cnt++;
  }
  return cnt;
}
void solve() {
  cin >> s;
  map<char, set<int> > mp;
  vector<vector<int> > sm;
  vector<int> tmp(26, 0);
  sm.push_back(tmp);
  for (int i = 0; i < s.length(); i++) {
    mp[s[i]].insert(i);
    sm.push_back(sm[sm.size() - 1]);
    sm[sm.size() - 1][s[i] - 'a']++;
  }
  tmp.assign(s.size(), 0);
  vector<vector<int> > palindroms(s.length(), tmp);
  for (int i = 0; i < s.length(); i++) {
    for (int j = i; j < s.length(); j++) {
      if (check(sm, i, j)) palindroms[i][j] = 1;
    }
  }
  long long int oldLeft = 0;
  long long int oldRight = 0;
  for (int i = 0; i < s.length(); i++) {
    for (int j = i; j < s.length(); j++) oldRight += palindroms[i][j];
  }
  long long int answ = 0;
  for (int i = 0; i < s.length() - 1; i++) {
    int d1 = 0, d2 = 0;
    for (int j = 0; j <= i; j++) {
      if (palindroms[j][i]) d1++;
    }
    for (int j = i; j < s.length(); j++) {
      if (palindroms[i][j]) d2++;
    }
    answ += d1 * (oldRight - d2);
    oldRight -= d2;
    oldLeft += d1;
  }
  cout << answ << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 0;
  if (t == 1)
    cin >> t;
  else
    t = 1;
  while (t--) solve();
}

```