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;
string s;
int ln;
vector<int> a[2005];
vector<int> b[2005];
long long before[2005] = {0};
long long after[2005] = {0};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  ln = (int)s.length();
  a[0].push_back(0);
  before[0] = 1;
  for (int i = 1; i < ln; i++) {
    a[i].push_back(i);
    for (int j = 0; j < (int)a[i - 1].size(); j++) {
      if (a[i - 1][j] > 0 && (s[i] == (s[a[i - 1][j] - 1])))
        a[i].push_back(a[i - 1][j] - 1);
    }
    if (s[i] == s[i - 1]) a[i].push_back(i - 1);
    before[i] = (int)a[i].size();
  }
  b[ln - 1].push_back(ln - 1);
  after[ln - 1] = 1;
  for (int i = ln - 2; i >= 0; i--) {
    b[i].push_back(i);
    for (int j = 0; j < (int)b[i + 1].size(); j++) {
      if (b[i + 1][j] < (ln - 1) && (s[i] == (s[b[i + 1][j] + 1])))
        b[i].push_back(b[i + 1][j] + 1);
    }
    if (s[i] == s[i + 1]) b[i].push_back(i + 1);
    after[i] = (int)b[i].size();
    after[i] += after[i + 1];
  }
  long long ans = 0;
  for (int i = 0; i < ln - 1; i++) ans += before[i] * after[i + 1];
  cout << ans << "\n";
}