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 main() {
  char a[2002];
  int b[2002] = {0};
  long long count = 0, n = 0;
  cin.getline(a, 2002);
  n = strlen(a);
  for (int i = 0; i < n - 1; i++) {
    long long count1 = 0, count2 = 0;
    for (int j = i - 1; j >= 0; j--) count1 += b[j];
    count2 = count1;
    if (i != 0) {
      int j = i - 1, k = i + 1;
      for (; j >= 0 && a[k] != '\0' && a[k] == a[j]; j--, k++) {
        b[k]++;
        count1 -= b[j];
        count += j + n - k - 1 + count1;
      }
      j++, k--;
    }
    int j = i, k = i + 1;
    count2 += b[j];
    for (; j >= 0 && a[k] != '\0' && a[k] == a[j]; j--, k++) {
      b[k]++;
      count2 -= b[j];
      count += j + n - k - 1 + count2;
    }
    j++, k--;
  }
  count += ((n * (n - 1)) / 2);
  cout << count << "\n";
}