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:

// ~/BAU/ACM-ICPC/Teams/A++/BlackBurn95
// ~/sudo apt-get Accpeted

import java.io.*;
import java.util.*;
import java.math.*;
import static java.lang.Math.*;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Double.parseDouble;
import static java.lang.String.*;

public class Main {
    
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder out = new StringBuilder();
        StringTokenizer tk;
        
        char [] s = in.readLine().toCharArray();
        int n = s.length;
        
        int [][] pal = new int[n][n];
        for(int i=0; i<n; i++)
            pal[i][i] = 1;
        for(int i=0; i<n-1; i++)
            pal[i][i+1] = (s[i]==s[i+1] ? 1 : 0);
        
        for(int l=3; l<=n; l++) {
            for(int i=0; i<=n-l; i++) {
                int j=i+l-1;
                pal[i][j] = (pal[i+1][j-1]==1 && s[i]==s[j] ? 1 : 0);
            }
        }
        
        long [] num = new long[n];
        long [] dp = new long[n];
        num[0] = 1;
        
        for(int i=1; i<n; i++) {
            dp[i] = dp[i-1];
            num[i] = num[i-1];
            for(int j=1; j<=i; j++) {
                dp[i] += pal[j][i]*num[j-1];
                num[i] += pal[j][i];
            }
            num[i] += pal[0][i];
        }
        /*
        long ans = 0;
        for(int i=0; i<n; i++)
            ans += dp[i];
                */
        System.out.println(dp[n-1]);
    }
    
}