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;
vector<int> plandromes[2010];
int isPland[2010][2010];
long long int dp[2010];
string in;
int solve(int start, int length) {
  if (length == 1 || length == 0) return 1;
  if (start >= in.size() || start < 0 || length < 0 ||
      start + length - 1 >= in.size())
    return false;
  if (isPland[start][length] != -1) return isPland[start][length];
  if (in[start] != in[start + length - 1])
    return isPland[start][length] = 0;
  else
    return isPland[start][length] = solve(start + 1, length - 2);
}
int main() {
  memset(isPland, -1, sizeof isPland);
  cin >> in;
  for (int i = 0; i < in.size(); i++) {
    for (int j = 0; j < in.size(); j++) {
      int is = solve(i, j + 1);
      if (is) {
        plandromes[i].push_back(j);
      }
    }
  }
  memset(dp, 0, sizeof dp);
  dp[in.size() - 1] = plandromes[in.size() - 1].size();
  for (int i = in.size() - 2; i >= 0; i--) {
    dp[i] = plandromes[i].size() + dp[i + 1];
  }
  long long int ans = 0;
  for (int i = 0; i < in.size(); i++) {
    for (int j = 0; j < plandromes[i].size(); j++) {
      ans += dp[i + plandromes[i][j] + 1];
    }
  }
  cout << ans << endl;
  return 0;
}