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 p[2014], q[2014];
char s[2014];
int main() {
  memset(p, 0, sizeof p);
  memset(q, 0, sizeof q);
  long long ans = 0;
  scanf("%s", s);
  int len = strlen(s);
  for (int i = 0; i < len; i++)
    for (int x = i, y = i; x >= 0 && y < len; x--, y++)
      if (s[x] == s[y])
        p[x]++, q[y]++;
      else
        break;
  for (int i = 1; i < len; i++)
    for (int x = i - 1, y = i; x >= 0 && y < len; x--, y++)
      if (s[x] == s[y])
        p[x]++, q[y]++;
      else
        break;
  for (int i = 0; i < len; i++)
    for (int j = i + 1; j < len; j++) ans += q[i] * p[j];
  printf("%I64d", ans);
  return 0;
}