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 JAVA):

Solution:
import java.util.*;
public class cf159d {
  static char[] v;
  static int n;
  static int[][] memo;
  public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    v = in.next().trim().toCharArray();
    int n = v.length;
    memo = new int[n][n];
    for(int[] x : memo) Arrays.fill(x, -1);
    int[] sum = new int[n+1];
    for(int i=0; i<n; i++)
      for(int j=i; j<n; j++)
        sum[i] += go(i,j);
    for(int i=n-1; i>=0; i--)
      sum[i] += sum[i+1];
    long ans = 0;
    for(int i=0; i<n; i++)
      for(int j=i; j<n; j++)
        ans += go(i,j)*1L*sum[j+1];
    System.out.println(ans);
  }
  static int go(int l, int r) {
    if(l > r) {
      return 1;
    }
    if(memo[l][r] != -1) return memo[l][r];
    int ans = 0;
    if(v[l]==v[r]) ans = go(l+1,r-1);
    return memo[l][r] = ans;
  }
}