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;
bool done[2004][2004];
bool dp[2004][2004];
long long ini[2004];
long long fin[2004];
string str;
int SZ;
bool ispali(int a, int b) {
  if (a == b) return dp[a][b] = true;
  if (abs(b - a) <= 1) return dp[a][b] = str[a] == str[b];
  if (done[a][b]) return dp[a][b];
  done[a][b] = true;
  bool &ans = dp[a][b];
  ans = ispali(a + 1, b - 1) && str[a] == str[b];
  ispali(a + 1, b);
  ispali(a, b - 1);
  return ans;
}
void Fini() {
  ini[0] = 1;
  for (int i = 1; i < SZ; ++i) {
    ini[i] = 0;
    for (int j = 0; j <= i; ++j) {
      ini[i] += ispali(j, i);
    }
  }
}
void Ffin() {
  fin[SZ - 1] = 1;
  for (int i = SZ - 2; i >= 0; --i) {
    fin[i] = 0;
    for (int j = SZ - 1; j >= i; --j) fin[i] += ispali(i, j);
  }
}
int main() {
  cin >> str;
  SZ = str.size();
  memset(done, 0, sizeof done);
  ispali(0, SZ - 1);
  Fini();
  Ffin();
  long long ans = 0;
  for (int i = 0; i < SZ; ++i)
    for (int j = i + 1; j < SZ; ++j) {
      ans += ini[i] * fin[j];
    }
  cout << ans << endl;
  return 0;
}