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;
vector<string> split(const string& s, char c) {
  vector<string> v;
  stringstream second(s);
  string x;
  while (getline(second, x, c)) v.emplace_back(x);
  return move(v);
}
template <typename T, typename... Args>
inline string arrStr(T arr, int n) {
  stringstream s;
  s << "[";
  for (int i = 0; i < n - 1; i++) s << arr[i] << ",";
  s << arr[n - 1] << "]";
  return s.str();
}
inline void __evars_begin(int line) { cerr << "#" << line << ": "; }
template <typename T>
inline void __evars_out_var(vector<T> val) {
  cerr << arrStr(val, val.size());
}
template <typename T>
inline void __evars_out_var(T* val) {
  cerr << arrStr(val, 10);
}
template <typename T>
inline void __evars_out_var(T val) {
  cerr << val;
}
inline void __evars(vector<string>::iterator it) { cerr << '\n'; }
template <typename T, typename... Args>
inline void __evars(vector<string>::iterator it, T a, Args... args) {
  cerr << it->substr((*it)[0] == ' ', it->length()) << "=";
  __evars_out_var(a);
  cerr << "; ";
  __evars(++it, args...);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int dx[8] = {1, -1, 0, 0, -1, -1, 1, 1};
int dy[8] = {0, 0, -1, 1, -1, 1, -1, 1};
const long long inf = 1000000007;
vector<int> vis(500010, 0), par(500010, 0), siz(500010, 0), len(500010, 0);
vector<int> adj[500010];
int myrandom(int i) { return std::rand() % i; }
const int MAX = 100100;
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s;
  cin >> s;
  int n = (int)s.length();
  long long dp[n + 1][n + 1];
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= n; i++) {
    dp[i][i] = 1;
    if (i < n && s[i - 1] == s[i]) dp[i][i + 1] = 1;
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= n - i; j++) {
      int st = j, en = j + i;
      if (dp[st + 1][en - 1] == 1 && s[st - 1] == s[en - 1]) dp[st][en] = 1;
    }
  }
  long long suf[n + 2];
  memset(suf, 0, sizeof(suf));
  for (int i = n; i >= 1; i--) {
    for (int j = i; j <= n; j++)
      if (dp[i][j] == 1) suf[i]++;
  }
  for (int i = n - 1; i >= 1; i--) suf[i] += suf[i + 1];
  long long tot = 0;
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++)
      if (dp[i][j] == 1) tot += suf[j + 1];
  cout << tot << '\n';
}