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 str;
long long ans[2001], sum[2001], temp, pal[2001][2001], n;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> str;
  n = str.size();
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n - i; j++) {
      int x = j;
      int y = j + i;
      if (x == y)
        pal[x][y] = 1;
      else if (x == y - 1 && str[x] == str[y])
        pal[x][y] = 1;
      else if (pal[x + 1][y - 1] == 1 && str[x] == str[y])
        pal[x][y] = 1;
    }
  for (int i = 0; i < n; i++) {
    temp = 0;
    for (int j = 0; j <= i; j++)
      if (pal[j][i]) temp++;
    sum[i] = (i ? sum[i - 1] : 0) + temp;
  }
  for (int i = 0; i < n; i++) {
    temp = 0;
    for (int j = 0; j <= i; j++) temp += pal[j][i] * sum[j - 1];
    ans[i] = ans[i - 1] + temp;
  }
  cout << ans[n - 1] << endl;
}