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;
long long dp[2005][2005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  long long n = s.size(), i, j, l;
  vector<pair<long long, long long> > v;
  for (i = 0; i < n; i++) dp[i][i] = 1;
  for (l = 2; l <= n; l++) {
    for (j = 0; j + l - 1 < n; j++) {
      i = j + l - 1;
      if (s[i] == s[j]) {
        if (i - j == 1)
          dp[j][i] = 1;
        else
          dp[j][i] = dp[j + 1][i - 1];
      }
    }
  }
  for (i = 0; i < n; i++) {
    for (j = i; j < n; j++) {
      if (dp[i][j]) {
        v.push_back(make_pair(i, j));
      }
    }
  }
  sort((v).begin(), (v).end());
  long long ans = 0;
  for (i = 0; i < v.size(); i++) {
    pair<long long, long long> p = v[i];
    long long idx =
        lower_bound((v).begin(), (v).end(), make_pair(p.second, 1000000007LL)) -
        v.begin();
    ans += v.size() - idx;
  }
  cout << ans;
  return 0;
}