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;
string str;
int pal[2005][2005];
long long palleft[2005], palright[2005];
int ispal(int x, int y) {
  if (pal[x][y] != -1) return pal[x][y];
  if (x == y) return (pal[x][y] = 1);
  if (x == y - 1) return (pal[x][y] = (str[x] == str[y]));
  if (str[x] != str[y]) return (pal[x][y] = 0);
  return (pal[x][y] = ispal(x + 1, y - 1));
}
int main() {
  memset(pal, -1, sizeof(pal));
  memset(palleft, 0, sizeof(palleft));
  memset(palright, 0, sizeof(palright));
  cin >> str;
  int len = ((int)(str).size());
  int i, j;
  for ((i) = 0; (i) < (int)(len); (i)++) {
    if (i > 0) palleft[i] += palleft[i - 1];
    for ((j) = 0; (j) < (int)(i + 1); (j)++) palleft[i] += ispal(j, i);
  }
  for (i = len - 1; i >= 0; i--) {
    if (i < len - 1) palright[i] += palright[i + 1];
    for (j = i; j < len; j++) palright[i] += ispal(i, j);
  }
  long long ans = 0;
  for ((i) = 0; (i) < (int)(len); (i)++) {
    for (j = i; j < len; j++) {
      if (!ispal(i, j)) continue;
      if (i > 0) ans += palleft[i - 1];
      if (j < len - 1) ans += palright[j + 1];
    }
  }
  ans /= 2;
  cout << ans << endl;
  cin.get();
  cin.get();
  return 0;
}