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 INF = 999999999;
vector<long long> H, Hr, step;
int n;
long long getHash(int l, int r) {
  long long h = H[r];
  if (l) h -= H[l - 1];
  return h * step[n - l];
}
long long getRevHash(int l, int r) {
  l = n - l - 1;
  r = n - r - 1;
  swap(l, r);
  long long h = Hr[r];
  if (l) h -= Hr[l - 1];
  return h * step[n - l];
}
int main() {
  int m, k, g, sum, num = 0, h, d;
  string s;
  cin >> s;
  n = s.size();
  vector<int> pcnt(n, 0), rcnt(n + 1, 0);
  H.resize(n);
  Hr.resize(n);
  step.resize(n + 1);
  long long x = 1;
  for (int(i) = 0; (i) < (n); ++(i)) {
    step[i] = x;
    if (i == 0) {
      H[0] = s[i];
      Hr[0] = s[n - i - 1];
    } else {
      H[i] = H[i - 1] + s[i] * x;
      Hr[i] = Hr[i - 1] + s[n - i - 1] * x;
    }
    x *= 31;
  }
  step[n] = x;
  for (int i = n - 1; i >= 0; i--) {
    for (int j = i; j < n; j++)
      if (getHash(i, j) == getRevHash(i, j)) pcnt[i]++;
    rcnt[i] = pcnt[i] + rcnt[i + 1];
  }
  long long ans = 0;
  for (int(i) = 0; (i) < (n); ++(i))
    for (int j = i; j < n; j++)
      if (getHash(i, j) == getRevHash(i, j)) {
        ans += rcnt[j + 1];
      }
  cout << ans;
  return 0;
}