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;
bool isPalin[2001][2001];
long long int A[2001];
long long int B[2001];
void init() {
  for (int i = 0; i < 2001; i++) {
    A[i] = 0;
    for (int j = 0; j < 2001; j++) isPalin[i][j] = false;
  }
}
string str;
int n;
int main() {
  cin >> str;
  n = str.length();
  init();
  for (int i = 0; i < n; i++) {
    isPalin[i][i] = true;
    for (int j = 1; j + i < n && i - j >= 0; j++) {
      if (str[i - j] != str[i + j]) break;
      isPalin[i - j][i + j] = true;
    }
    for (int j = 0; j + i + 1 < n && i - j >= 0; j++) {
      if (str[i - j] != str[i + j + 1]) break;
      isPalin[i - j][i + j + 1] = true;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++)
      if (isPalin[j][i]) A[i]++;
    if (i) A[i] += A[i - 1];
  }
  long long ans = 0;
  for (int i = 1; i < n; i++)
    for (int j = i; j < n; j++)
      if (isPalin[i][j]) {
        ans += A[i - 1];
      }
  cout << ans << endl;
  return 0;
}