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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;


public class CF_159D {
    private static String line;
    private static int[][] dp = new int[2005][2005];
    private static int[] dp2 = new int[2005];
    
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        
        line = br.readLine();
        for (int i = 0; i < dp.length; i++)
            for (int j = 0; j < dp[i].length; j++)
                dp[i][j] = -1;
        
        for (int i = 0; i < line.length(); i++)
            for (int j = i; j < line.length(); j++)
                dp2[i] += f(i, j);
        
        for (int i = 1; i < dp2.length; i++)
            dp2[i] += dp2[i - 1];
        
        long ans = 0;
        
        for (int i = 0; i < line.length(); i++)
            for (int j = i; j < line.length() - 1; j++)
                if (f(i, j) == 1)
                    ans += dp2[line.length()] - dp2[j];
        
        out.println(ans);
        
        br.close();
        out.close();
    }
    
    private static int f(int i, int j) {
        if (dp[i][j] != -1)
            return dp[i][j];
        
        if (i == j)
            return dp[i][j] = 1;
        
        if (line.charAt(i) != line.charAt(j))
            return dp[i][j] = 0;
        
        if (j - i == 1)
            return dp[i][j] = 1;
        
        return dp[i][j] = f(i + 1, j - 1);
    }
}