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[2500];
int n;
long long res, a[2500];
bool pal[2500][2500];
void innit() {
  n = strlen(s);
  int i, j;
  for (i = n; i >= 1; i--) s[i] = s[i - 1];
  s[n + 1] = '\n';
  for (i = n; i >= 1; i--)
    for (j = i; j <= n; j++)
      if ((s[i] == s[j]) && ((i + 1 >= j) || (pal[i + 1][j - 1] == 1)))
        pal[i][j] = 1;
}
void tinh_a() {
  a[0] = 0;
  int i, j;
  for (i = 1; i <= n; i++) {
    a[i] = a[i - 1];
    for (j = i; j >= 1; j--)
      if (pal[j][i]) a[i]++;
  }
}
void tinh_kq() {
  res = 0;
  int i, j;
  long long t;
  for (i = n; i >= 2; i--) {
    t = 0;
    for (j = i; j <= n; j++) t += pal[i][j];
    res += t * a[i - 1];
  }
}
void solve() {
  innit();
  tinh_a();
  tinh_kq();
}
int main() {
  gets(s);
  solve();
  cout << res;
}