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;
char s[4000];
long long l[4000];
long long par_sum[4000];
int main(void) {
  scanf("%s", &s);
  string a = (string)s;
  memset(l, 0, sizeof(l));
  for (int i = 0; i < a.length(); i++) {
    for (int j = 0; j < a.length(); j++) {
      if (((i - j) >= 0) && ((i + j) < a.length())) {
        if (a[i - j] == a[i + j]) {
          l[i - j]++;
        } else
          break;
      } else
        break;
    }
    for (int j = 0; j < a.length(); j++) {
      if (((i - j) >= 0) && ((i + j + 1) < a.length())) {
        if (a[i - j] == a[i + j + 1])
          l[i - j]++;
        else
          break;
      } else
        break;
    }
  }
  par_sum[a.length()] = 0;
  for (int i = a.length() - 1; i >= 0; i--) par_sum[i] = par_sum[i + 1] + l[i];
  long long ans = 0;
  for (int i = 0; i < a.length(); i++) {
    for (int j = 0; j < a.length(); j++) {
      if (((i - j) >= 0) && ((i + j) < a.length())) {
        if (a[i - j] == a[i + j]) {
          ans += par_sum[i + j + 1];
        } else
          break;
      } else
        break;
    }
    for (int j = 0; j < a.length(); j++) {
      if (((i - j) >= 0) && ((i + j + 1) < a.length())) {
        if (a[i - j] == a[i + j + 1])
          ans += par_sum[i + j + 2];
        else
          break;
      } else
        break;
    }
  }
  printf("%I64d\n", ans);
  return 0;
}