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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC optimize("no-stack-protector,fast-math")
using namespace std;
const int N = 2e3 + 5, M = 1e4 + 5, MOD = 1e9 + 7;
string s;
long long ans;
int dp[N][N], n, pref[N][N];
void bulid() {
  for (int st = 0; st <= n; st++) dp[st][0] = dp[st][1] = 1;
  for (int sz = 2; sz <= n; sz++) {
    for (int st = 0; sz + st - 1 < n; st++) {
      int &ret = dp[st][sz];
      ret = (s[st] == s[sz + st - 1]) && dp[st + 1][sz - 2];
    }
  }
  for (int st = 0; st < n; st++)
    for (int sz = 1; sz <= n; sz++) dp[st][sz] += dp[st][sz - 1];
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  cin >> s;
  n = s.size();
  bulid();
  for (int i = 0; i < n; i++) {
    long long r = 0;
    for (int st = 0; st <= i; st++) {
      r += dp[st][i - st + 1] - dp[st][i - st];
    }
    long long l = 0;
    for (int st = i + 1; st < n; st++) l += dp[st][n - st] - dp[st][0];
    ans += l * r;
  }
  cout << ans << "\n";
  return 0;
}