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;
int main() {
  string S;
  cin >> S;
  int size = S.size();
  vector<long long int> startI(size, 0LL);
  vector<long long int> endJ(size, 0LL);
  for (int i = (0); i < (size); ++i) {
    endJ[i] += 1LL;
    startI[i] += 1LL;
    for (int j = (1); j < (size); ++j) {
      if (i - j >= 0 && i + j < size && S[i - j] == S[i + j] &&
          !((i - j == 0) && (i + j == size - 1))) {
        endJ[i + j] += 1LL;
        startI[i - j] += 1LL;
      } else
        break;
    }
  }
  for (int i = (0); i < (size); ++i) {
    if (S[i] == S[i + 1]) {
      startI[i] += 1LL;
      endJ[i + 1] += 1LL;
      for (int j = (1); j < (size); ++j) {
        if (i - j >= 0 && i + j < size && S[i - j] == S[i + 1 + j]) {
          endJ[i + j + 1] += 1LL;
          startI[i - j] += 1LL;
        } else
          break;
      }
    }
  }
  long long int total = 0LL;
  for (int i = size - 1; i >= 1; --i) startI[i - 1] += startI[i];
  for (int i = 0; i < size - 1; ++i) total += endJ[i] * startI[i + 1];
  cout << total << endl;
}