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;
const int INF = 1000 * 1000 * 1001;
int isp[2010][2010];
string s;
int n;
int Isp(int i, int j) {
  if (isp[i][j] == -1) {
    if (j - i == 0)
      isp[i][j] = 1;
    else if (j - i == 1) {
      if (s[i] == s[j])
        isp[i][j] = 1;
      else
        isp[i][j] = 0;
    } else
      isp[i][j] = s[i] == s[j] ? Isp(i + 1, j - 1) : 0;
  }
  return isp[i][j];
}
int main() {
  cin >> s;
  memset(isp, -1, sizeof isp);
  n = s.size();
  vector<long long> dp1(n), dp2(n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= i; j++) dp1[i] += Isp(j, i);
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++) dp2[i] += Isp(i, j);
  long long res = 0;
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) res += dp1[i] * dp2[j];
  cout << res << endl;
  return 0;
}