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;
long long x[2005], dp[2005][2005];
string s;
bool rec(long long a, long long b) {
  if (b < a) {
    return 1;
  }
  long long &re = dp[a][b];
  if (re != -1) {
    return re;
  }
  if (s[a] == s[b]) {
    return re = rec(a + 1, b - 1);
  } else {
    return re = 0;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  ;
  cin >> s;
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < s.size(); i++) {
    string ss;
    for (int j = i; j < s.size(); j++) {
      ss += s[j];
      if (rec(i, j)) {
        x[i]++;
      }
    }
  }
  for (int i = 1; i < 2005; i++) {
    x[i] += x[i - 1];
  }
  long long sum = x[s.size() - 1], n = 0;
  for (int i = 0; i < s.size(); i++) {
    string ss;
    for (int j = i; j < s.size(); j++) {
      ss += s[j];
      if (rec(i, j)) {
        n += abs(sum - x[j]);
      }
    }
  }
  cout << n << endl;
  return 0;
}