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;
const int N = 4024;
int a[N], n, an;
int f[N];
long long g[N], res;
char s[N];
inline bool check(int x, int y) {
  if (x < 0) return 0;
  if (y > n) return 0;
  if (a[x] != a[y]) return 0;
  return 1;
}
int main() {
  scanf("%s", s);
  n = strlen(s);
  for (int i = 1; i <= n; i++) {
    a[++an] = s[i - 1] - 'a' + 1;
    a[++an] = 0;
  }
  --an;
  n = an;
  for (int i = 1; i <= n; i++) {
    f[i] = 1;
    while (check(i - f[i], i + f[i])) ++f[i];
  }
  res = 0;
  for (int i = 1; i <= n; i++)
    for (int j = (i & 1) ? i : i + 1; j <= i + f[i] - 1; j += 2) g[j]++;
  g[0] = 0;
  for (int i = 1; i <= n; i++) g[i] += g[i - 1];
  for (int i = 1; i <= n; i++)
    for (int j = (i & 1) ? i : i - 1; j >= i - f[i] + 1; j -= 2)
      res += g[j - 1];
  cout << res << endl;
  return 0;
}