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>
int tab[4096];
unsigned long long res = 0;
int main() {
  for (int i = 0; i < 4096; i++) tab[i] = 0;
  std::string str;
  std::cin >> str;
  int s = str.size();
  for (int i = 0; i < s; i++) {
    for (int j = 0; j <= i && j + i < s; j++)
      if (str[i + j] != str[i - j])
        break;
      else {
        int p = 0, k = 2048, id = 0;
        while (k - p > 1) {
          int t = (k + p) / 2;
          if (t > i + j) {
            tab[id * 2 + 2]++;
            id = id * 2 + 1;
            k = t;
          } else {
            id = id * 2 + 2;
            p = t;
          }
        }
        p = 0;
        k = 2048;
        id = 0;
        while (k - p > 1) {
          int t = (k + p) / 2;
          if (t > i - j) {
            id = id * 2 + 1;
            res += tab[id];
            k = t;
          } else {
            id = id * 2 + 2;
            res += tab[id];
            p = t;
          }
        }
      }
    for (int j = 0; j < i; j++)
      if (str[i - j - 1] != str[i + j])
        break;
      else {
        int p = 0, k = 2048, id = 0;
        while (k - p > 1) {
          int t = (k + p) / 2;
          if (t > i + j) {
            tab[id * 2 + 2]++;
            id = id * 2 + 1;
            k = t;
          } else {
            id = id * 2 + 2;
            p = t;
          }
        }
        p = 0;
        k = 2048;
        id = 0;
        while (k - p > 1) {
          int t = (k + p) / 2;
          if (t > i - j - 1) {
            id = id * 2 + 1;
            res += tab[id];
            k = t;
          } else {
            id = id * 2 + 2;
            res += tab[id];
            p = t;
          }
        }
      }
  }
  std::cout << res;
  return 0;
}