#include <bits/stdc++.h>
using namespace std;
const long long X = 137LL, X2 = 139LL;
const long long M1 = 1000000007LL;
const long long M2 = 1000000009LL;
const int N = 2005;
long long h[N];
long long r[N];
long long rh[N];
long long rr[N];
void buildHash(string s) {
  h[0] = rh[0] = 0;
  r[0] = rr[0] = 1;
  int n = s.length();
  for (int i = 1; i <= n; i++) {
    r[i] = (((r[i - 1] * X) % M1 + M1) % M1);
    h[i] = (((h[i - 1] * X + s[i - 1]) % M1 + M1) % M1);
  }
  reverse(s.begin(), s.end());
  for (int i = 1; i <= n; i++) {
    rr[i] = (((rr[i - 1] * X) % M1 + M1) % M1);
    rh[i] = (((rh[i - 1] * X + s[i - 1]) % M1 + M1) % M1);
  }
}
int n;
bool isPalindrome(int i, int j) {
  i++;
  j++;
  if (i == j) return true;
  int ni, nj;
  ni = n - j + 1;
  nj = n - i + 1;
  long long norm =
      (((h[j] - (((h[i - 1] * r[j - i + 1]) % M1 + M1) % M1)) % M1 + M1) % M1);
  long long reve =
      (((rh[nj] - (((rh[ni - 1] * rr[nj - ni + 1]) % M1 + M1) % M1)) % M1 +
        M1) %
       M1);
  return norm == reve;
}
vector<pair<int, int> > pal;
long long cont[N];
long long cont2[N];
int main() {
  string s;
  cin >> s;
  n = s.length();
  buildHash(s);
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 0; j <= i; j++) {
      if (isPalindrome(j, i)) {
        pal.push_back(make_pair(j, i));
        cont[j]++;
      }
    }
  }
  cont[n] = 0;
  for (int i = n - 1; i >= 0; i--) {
    cont[i] += cont[i + 1];
  }
  long long res = 0LL;
  for (int i = 0; i < pal.size(); i++) {
    res += cont[pal[i].second + 1];
  }
  cout << res << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 long long X = 137LL, X2 = 139LL;
const long long M1 = 1000000007LL;
const long long M2 = 1000000009LL;
const int N = 2005;
long long h[N];
long long r[N];
long long rh[N];
long long rr[N];
void buildHash(string s) {
  h[0] = rh[0] = 0;
  r[0] = rr[0] = 1;
  int n = s.length();
  for (int i = 1; i <= n; i++) {
    r[i] = (((r[i - 1] * X) % M1 + M1) % M1);
    h[i] = (((h[i - 1] * X + s[i - 1]) % M1 + M1) % M1);
  }
  reverse(s.begin(), s.end());
  for (int i = 1; i <= n; i++) {
    rr[i] = (((rr[i - 1] * X) % M1 + M1) % M1);
    rh[i] = (((rh[i - 1] * X + s[i - 1]) % M1 + M1) % M1);
  }
}
int n;
bool isPalindrome(int i, int j) {
  i++;
  j++;
  if (i == j) return true;
  int ni, nj;
  ni = n - j + 1;
  nj = n - i + 1;
  long long norm =
      (((h[j] - (((h[i - 1] * r[j - i + 1]) % M1 + M1) % M1)) % M1 + M1) % M1);
  long long reve =
      (((rh[nj] - (((rh[ni - 1] * rr[nj - ni + 1]) % M1 + M1) % M1)) % M1 +
        M1) %
       M1);
  return norm == reve;
}
vector<pair<int, int> > pal;
long long cont[N];
long long cont2[N];
int main() {
  string s;
  cin >> s;
  n = s.length();
  buildHash(s);
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 0; j <= i; j++) {
      if (isPalindrome(j, i)) {
        pal.push_back(make_pair(j, i));
        cont[j]++;
      }
    }
  }
  cont[n] = 0;
  for (int i = n - 1; i >= 0; i--) {
    cont[i] += cont[i + 1];
  }
  long long res = 0LL;
  for (int i = 0; i < pal.size(); i++) {
    res += cont[pal[i].second + 1];
  }
  cout << res << endl;
  return 0;
}

```