#include <bits/stdc++.h>
using namespace std;
int n;
char str[2002];
int palind[2002][2002];
int solve(int i, int j) {
  if (i >= j) return true;
  int &ret = palind[i][j];
  if (ret != -1) return ret;
  ret = false;
  if (str[i] == str[j]) ret = solve(i + 1, j - 1);
  return ret;
}
int main() {
  scanf(" %s", str);
  n = strlen(str);
  memset(palind, -1, sizeof palind);
  long long ans = 0;
  vector<pair<long long, long long>> v;
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++)
      if (solve(i, j)) v.push_back(make_pair(i, j));
  sort(v.begin(), v.end());
  for (int i = 0; i < (int)v.size(); i++) {
    long long startBeforeThisEnd = v[i].second + 1;
    startBeforeThisEnd =
        upper_bound(v.begin(), v.end(), make_pair(startBeforeThisEnd, -1LL)) -
        v.begin();
    ans += (int)v.size() - startBeforeThisEnd;
  }
  cout << ans << endl;
}

### 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;
int n;
char str[2002];
int palind[2002][2002];
int solve(int i, int j) {
  if (i >= j) return true;
  int &ret = palind[i][j];
  if (ret != -1) return ret;
  ret = false;
  if (str[i] == str[j]) ret = solve(i + 1, j - 1);
  return ret;
}
int main() {
  scanf(" %s", str);
  n = strlen(str);
  memset(palind, -1, sizeof palind);
  long long ans = 0;
  vector<pair<long long, long long>> v;
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++)
      if (solve(i, j)) v.push_back(make_pair(i, j));
  sort(v.begin(), v.end());
  for (int i = 0; i < (int)v.size(); i++) {
    long long startBeforeThisEnd = v[i].second + 1;
    startBeforeThisEnd =
        upper_bound(v.begin(), v.end(), make_pair(startBeforeThisEnd, -1LL)) -
        v.begin();
    ans += (int)v.size() - startBeforeThisEnd;
  }
  cout << ans << endl;
}

```