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;
using namespace std;
long long power(long long b, long long e, long long m) {
  long long p = 1;
  while (e > 0) {
    if (e & 1) p = (p * b) % m;
    e = e >> 1;
    b = (b * b) % m;
  }
  return p;
}
string s;
bool isPal[2001][2001];
long long psh[2001], peh[2001];
int main() {
  long long n, i, t, j, h, k = 0, h1, h2, q, l, r, ans = 0;
  cin >> s;
  n = s.length();
  isPal[n - 1][n - 1] = true;
  for (i = 0; i < n - 1; i++) {
    isPal[i][i] = true;
    if (s[i] == s[i + 1]) {
      isPal[i][i + 1] = true;
    }
  }
  for (i = 2; i < n; i++) {
    for (j = 0; j + i < n; j++) {
      isPal[j][j + i] = isPal[j + 1][j + i - 1] && (s[j] == s[j + i]);
    }
  }
  peh[0] = psh[n - 1] = 1;
  for (i = 1; i < n; i++) {
    peh[i] = peh[i - 1];
    for (j = 0; j <= i; j++) {
      peh[i] += isPal[j][i];
    }
  }
  for (i = n - 2; i >= 0; i--) {
    for (j = i; j < n; j++) {
      psh[i] += isPal[i][j];
    }
  }
  for (i = 0; i < n - 1; i++) {
    ans += peh[i] * psh[i + 1];
  }
  printf("%lld\n", ans);
  return 0;
}