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 maxn = 2010, r = 19;
int z[maxn], z2[maxn], t[maxn];
vector<int> sar, tah;
int main() {
  t[0] = 1;
  for (int i = 1; i < maxn; ++i) t[i] = t[i - 1] * r;
  string s;
  cin >> s;
  z[0] = int(s[0]);
  z2[s.size() - 1] = int(s[s.size() - 1]);
  for (int i = 1; i < s.size(); ++i) z[i] = z[i - 1] * r + int(s[i]);
  for (int i = s.size() - 2; i >= 0; i--) z2[i] = z2[i + 1] * r + int(s[i]);
  for (int i = 0; i < s.size(); ++i)
    for (int j = i; j < s.size(); ++j)
      if (z[j] - (i ? z[i - 1] * t[j - i + 1] : 0) ==
          z2[i] - (j != s.size() - 1 ? z2[j + 1] * t[j - i + 1] : 0)) {
        sar.push_back(i);
        tah.push_back(j);
      }
  sort(sar.begin(), sar.end());
  sort(tah.begin(), tah.end());
  long long ans = 0;
  for (int i = 0; i < tah.size(); ++i)
    ans += sar.size() -
           int(upper_bound(sar.begin(), sar.end(), tah[i]) - sar.begin());
  cout << ans << endl;
  return 0;
}