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 N;
string S;
long long unsigned A[5000], B[5000];
long long unsigned A2[5000], B2[5000];
int main() {
  int yx, yy;
  long long unsigned int i;
  cin >> S;
  N = S.size();
  int bas, sol, sag;
  for (i = 0; i < N; i++) {
    sol = sag = i;
    while (sol >= 0 && sag < N && sol <= sag && S[sol] == S[sag]) {
      A[sol + 1]++;
      B[sag + 1]++;
      sol--;
      sag++;
    }
    sol = sag = i;
    sag++;
    while (sol >= 0 && sag < N && sol <= sag && S[sol] == S[sag]) {
      A[sol + 1]++;
      B[sag + 1]++;
      sol--;
      sag++;
    }
  }
  for (i = 1; i <= N; i++) A[i] += A[i - 1];
  for (i = 1; i <= N; i++) B[i] += B[i - 1];
  long long unsigned int cvp = 0;
  for (i = 1; i <= N; i++) cvp += (B[i] - B[i - 1]) * (A[N] - A[i]);
  cout << cvp << endl;
  return 0;
}