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;
const int mx = 2e3 + 4;
int mark[mx][mx], dppa[mx], dppk[mx];
int main() {
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); i++) mark[i][i] = 1, mark[i + 1][i] = 1;
  for (int j = 1; j < s.size(); j++)
    for (int i = 0; i + j < s.size(); i++)
      if (s[j + i] == s[i] && mark[i + 1][j + i - 1]) mark[i][i + j] = 1;
  for (int i = 0; i < s.size(); i++)
    for (int j = i; j < s.size(); j++)
      dppa[i] += mark[i][j], dppk[j] += mark[i][j];
  for (int i = 1; i < s.size(); i++) dppk[i] += dppk[i - 1];
  for (int i = s.size() - 2; i > -1; i--) dppa[i] += dppa[i + 1];
  long long ans = 0;
  for (int i = 0; i < s.size(); i++) {
    for (int j = i; j < s.size(); j++) {
      if (mark[i][j]) {
        ans += dppa[j + 1];
        if (i != 0) ans += dppk[i - 1];
      }
    }
  }
  cout << ans / 2;
}