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 Ma[4105];
int Mp[4105];
void Manacher(char s[], int len) {
  int l = 0;
  Ma[l++] = '$';
  Ma[l++] = '#';
  for (int i = 0; i < len; i++) {
    Ma[l++] = s[i];
    Ma[l++] = '#';
  }
  Ma[l] = 0;
  int mx = 0, id = 0;
  for (int i = 0; i < l; i++) {
    Mp[i] = mx > i ? min(Mp[id * 2 - i], mx - i) : 1;
    while (Ma[i + Mp[i]] == Ma[i - Mp[i]]) Mp[i]++;
    if (i + Mp[i] > mx) {
      mx = i + Mp[i];
      id = i;
    }
  }
}
char s[4105];
long long f[4105];
int ans, cnt;
int tmp[2005 * 2005];
int main() {
  scanf("%s", s);
  int len = strlen(s);
  Manacher(s, len);
  long long ans = 0;
  for (int i = 1; i < 2 * len + 2; i++) {
    for (int j = 1; j < Mp[i]; j++)
      if (Ma[i - j + 1] != '#') {
        tmp[++cnt] = i + j - 1;
        f[i - j + 1]++;
      }
  }
  for (int i = 2 * len + 2; ~i; i--) f[i] += f[i + 1];
  for (int i = 1; i <= cnt; i++) ans += f[tmp[i] + 1];
  printf("%I64d\n", ans);
}