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 MAXN = 2e3 + 10, inf = 1e9;
void make(string s, vector<int> dp[]) {
  dp[0].push_back(0);
  for (int i = 1; i < s.length(); i++) {
    vector<int> v = dp[i - 1], ans;
    if (s[i] == s[i - 1]) ans.push_back(i - 1);
    for (int j = 0; j < v.size(); j++) {
      int x = v[j] - 1;
      if (x >= 0 and s[x] == s[i]) ans.push_back(x);
    }
    ans.push_back(i);
    dp[i] = ans;
  }
}
vector<int> dp1[MAXN], dp2[MAXN];
long long d1[MAXN], d2[MAXN];
int main() {
  string s;
  cin >> s;
  make(s, dp1);
  reverse(s.begin(), s.end());
  make(s, dp2);
  reverse(dp2, dp2 + s.length());
  d1[0] = dp1[0].size();
  for (int i = 1; i < s.length(); i++) {
    d1[i] = d1[i - 1] + dp1[i].size();
  }
  d2[s.length() - 1] = dp2[s.length() - 1].size();
  for (int i = s.length() - 2; i >= 0; i--) {
    d2[i] = d2[i + 1] + dp2[i].size();
  }
  long long ans = 0;
  for (int i = 0; i < s.length() - 1; i++) ans += dp1[i].size() * d2[i + 1];
  cout << ans << endl;
  return 0;
}