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;
bool f[2001][2001];
long long fs[2001];
string s;
int n;
void update(int i) {
  i++;
  while (i <= n) {
    fs[i]++;
    i += i & -i;
  }
}
long long get(int i) {
  i++;
  long long ret = 0;
  while (i > 0) {
    ret += fs[i];
    i -= i & -i;
  }
  return ret;
}
int main() {
  cin >> s;
  n = s.length();
  memset(f, false, sizeof f);
  memset(fs, 0, sizeof fs);
  for (int i = 0; i <= 1; i++)
    for (int j = 0; j < n; j++) f[i][j] = true;
  for (int i = 2; i <= n; i++)
    for (int j = i - 1; j < n; j++)
      f[i][j] = f[i - 2][j - 1] && (s[j] == s[j - i + 1]);
  long long result = 0;
  for (int j = 0; j < n; j++)
    for (int i = 1; i <= j + 1; i++)
      if (f[i][j]) {
        result += get(j - i);
        update(j);
      }
  cout << result;
  return 0;
}