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 lst[2011][2011];
long long next_starts[2011];
long long strt_point[2011];
char str[3001];
int main() {
  gets(str);
  memset(next_starts, 0, sizeof(next_starts));
  memset(strt_point, 0, sizeof(strt_point));
  int len = strlen(str);
  register int i, j;
  for (i = 0; i < len; i++) {
    lst[i][strt_point[i]++] = i;
    for (j = i + 1; j < len; j++) {
      if (str[i] != str[j]) continue;
      int s, e;
      s = i;
      e = j;
      bool pal = true;
      while (s <= e) {
        if (str[s] != str[e]) {
          pal = false;
          break;
        }
        s++;
        e--;
      }
      if (pal) {
        lst[i][strt_point[i]++] = j;
      }
    }
  }
  next_starts[len - 1] = strt_point[len - 1];
  for (i = len - 2; i >= 0; i--)
    next_starts[i] = strt_point[i] + next_starts[i + 1];
  long long count = 0;
  for (i = 0; i < len; i++) {
    for (j = 0; j < strt_point[i]; j++) {
      int st = i;
      int en = lst[i][j];
      count += next_starts[en + 1];
    }
  }
  cout << count << endl;
  return 0;
}