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;
struct pairs {
  int i, j;
};
bool ispal(string& str, int i, int j) {
  while (i < j) {
    if (str[i++] != str[j--]) {
      return false;
    }
  }
  return true;
}
int main() {
  string str;
  cin >> str;
  pairs tmp;
  vector<pairs> v;
  vector<long long> counter;
  long long num;
  for (int i = 0; i < str.length(); i++) {
    num = 0;
    for (int j = i; j < str.length(); j++) {
      if (ispal(str, i, j)) {
        tmp.i = i;
        tmp.j = j;
        v.push_back(tmp);
        num++;
      }
    }
    counter.push_back(num);
  }
  long long cnt(0);
  for (int i = 0; i < v.size(); i++) {
    for (int j = v[i].j + 1; j < str.length(); j++) {
      cnt += counter[j];
    }
  }
  cout << cnt << endl;
  return 0;
}