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 long long N = 2e3 + 2;
char s[N];
long long ans[N], revans[N];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> s + 1;
  long long n = strlen(s + 1);
  for (long long ind = 1; ind <= n; ++ind) {
    long long i = ind - 1, j = ind;
    while (i && j <= n) {
      if (s[i] != s[j]) break;
      ++ans[j];
      ++j, --i;
    }
  }
  for (long long ind = 1; ind <= n; ++ind) {
    long long i = ind, j = ind;
    while (i && j <= n) {
      if (s[i] != s[j]) break;
      ++ans[j];
      ++j, --i;
    }
  }
  for (long long ind = n; ind; --ind) {
    long long i = ind - 1, j = ind;
    while (i && j <= n) {
      if (s[i] != s[j]) break;
      ++revans[i];
      ++j, --i;
    }
  }
  for (long long ind = n; ind; --ind) {
    long long i = ind, j = ind;
    while (i && j <= n) {
      if (s[i] != s[j]) break;
      ++revans[i];
      ++j, --i;
    }
  }
  long long sum = 0;
  for (long long i = 1; i <= n; ++i)
    for (long long j = i + 1; j <= n; ++j) sum += ans[i] * revans[j];
  cout << sum;
  return 0;
}