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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << "\t";
  err(++it, args...);
}
const int MOD = 1000000007;
const int N = 2e5;
long long int mpow(long long int base, long long int exp) {
  base %= MOD;
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((long long int)result * base) % MOD;
    base = ((long long int)base * base) % MOD;
    exp >>= 1;
  }
  return result;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string second;
  cin >> second;
  int n = second.size();
  bool palin[n][n];
  for (int i = (n - 1); n - 1 < -1 ? i < -1 : i > -1; n - 1 < -1 ? i++ : i--) {
    for (int j = (0); 0 < n ? j < n : j > n; 0 < n ? j++ : j--) {
      if (i >= j)
        palin[i][j] = true;
      else {
        palin[i][j] = (second[i] == second[j] && palin[i + 1][j - 1]);
      }
    }
  }
  int prefix[n + 1];
  memset(prefix, 0, sizeof(prefix));
  long long int ans = 0;
  for (int i = (0); 0 < n ? i < n : i > n; 0 < n ? i++ : i--) {
    int len = 0;
    for (int j = (0); 0 < i + 1 ? j < i + 1 : j > i + 1; 0 < i + 1 ? j++ : j--)
      if (palin[j][i]) {
        ans += prefix[j];
        len++;
      }
    prefix[i + 1] = prefix[i] + len;
  }
  cout << ans;
  return 0;
}