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.IOException;
import java.io.InputStreamReader;


public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
        
        char[] a = r.readLine().toCharArray();
        int N = a.length;
        
        boolean[][] P = new boolean[N + 10][N + 10];
        for(int i = 0; i < N; i++)
            P[i][i] = true;
        
        for(int l = 2; l <= N; l++)
        for(int i = 0; i <= N - l; i++){
            int j = i + l - 1;
            
            if(a[i] == a[j]){
                if(l == 2)P[i][j] = true;
                else P[i][j] = P[i + 1][j - 1];
            }else P[i][j] = false;
        }
        
        long[] c = new long[N + 10];
        long[] R = new long[N + 10];
        
        for(int i = 0; i < N; i++)
        for(int j = i; j < N; j++)
            if(P[i][j]) c[i]++;
        
        R[N - 1] = c[N - 1];
        for(int i = N - 2; i >= 0; i--)
            R[i] = c[i] + R[i + 1];
        
        long sum = 0;
        for(int i = 0; i < N; i++)
        for(int j = i; j < N; j++)if(P[i][j])
            sum += R[j + 1];
        
        System.out.println(sum);
    }
}