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 INF = 2147483647;
const int MaxS = 3000;
string str;
int l, r, n;
long long d1[MaxS], d2[MaxS], ans;
int main() {
  getline(cin, str);
  for (int i = 0; i < str.size(); i++) {
    l = i;
    r = str.size() - i - 1;
    n = min(l, r);
    for (int j = 0; j <= n; j++)
      if (str[i - j] != str[i + j])
        break;
      else {
        d1[i + j]++;
        d2[i - j]++;
      }
    l = i;
    r = str.size() - i;
    n = min(l, r) - 1;
    for (int j = 0; j <= n; j++)
      if (str[i - j - 1] != str[i + j])
        break;
      else {
        d1[i + j]++;
        d2[i - j - 1]++;
      }
  }
  for (int i = str.size() - 1; i >= 0; i--) d2[i] += d2[i + 1];
  for (int i = 0; i < str.size() - 1; i++) ans += d1[i] * d2[i + 1];
  printf("%I64d\n", ans);
}