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 long long INF = 1000000000 + 7;
const double esp = 1e-13;
const double pi = 3.141592653589;
long long ans, d[2000 + 10];
bool f[2000 + 10][2000 + 10];
int n, j;
char s[2000 + 10];
int main() {
  cin >> (s + 1);
  memset(f, false, sizeof(f));
  n = strlen(s + 1);
  for (int i = (1), _b = (n); i <= _b; i++) {
    f[i][i] = true;
    if (s[i] == s[i + 1]) {
      f[i][i + 1] = true;
      f[i + 1][i] = true;
    }
  }
  for (int k = (1), _b = (n); k <= _b; k++)
    for (int i = (1), _b = (n - k); i <= _b; i++) {
      j = i + k;
      if (f[i + 1][j - 1] && s[i] == s[j]) {
        f[i][j] = true;
        f[j][i] = true;
      }
    }
  ans = 0;
  for (int i = (1), _b = (n); i <= _b; i++)
    for (int j = (i), _b = (n); j <= _b; j++)
      if (f[i][j]) d[j]++;
  for (int i = (1), _b = (n); i <= _b; i++) d[i] += d[i - 1];
  for (int i = (2), _b = (n); i <= _b; i++)
    for (int j = (i), _b = (n); j <= _b; j++)
      if (f[i][j]) ans += d[i - 1];
  cout << ans;
}