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;
int n;
string s;
vector<long long> count_start;
vector<long long> count_finish;
void count_polyndroms(int pos) {
  int len = 0;
  while (pos - len >= 0 && pos + len < n) {
    if (s[pos - len] != s[pos + len]) break;
    count_start[pos - len]++;
    count_finish[pos + len]++;
    ++len;
  }
  len = 1;
  while (pos - len + 1 >= 0 && pos + len < n) {
    if (s[pos - len + 1] != s[pos + len]) break;
    count_start[pos - len + 1]++;
    count_finish[pos + len]++;
    ++len;
  }
}
int main() {
  cin >> s;
  n = s.size();
  count_finish.resize(n, 0);
  count_start.resize(n, 0);
  for (int i = 0; i < n; ++i) {
    count_polyndroms(i);
  }
  long long res = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      res += count_start[j] * count_finish[i];
    }
  }
  cout << res;
}