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.Scanner;

public class PalindromePairs {
    
    private final char[] chars;
    private int[][] palindrome;
    
    private PalindromePairs(String input) {
        chars = input.toCharArray();
    }
    
    
    private long solve() {
        
        int N = chars.length;
        
        boolean[][] DP = new boolean[N][N];
        
        for (int S = 0; S < N; S++) DP[S][S] = true;
        for (int S = 0; S < N - 1; S++) DP[S][S + 1] = chars[S] == chars[S + 1];
        
        for (int L = 2; L < N; L++) {
            for (int S = 0; S < N - L; S++) 
                DP[S][S + L] = (chars[S] == chars[S + L] ? DP[S + 1][S + L - 1] : false);
        }
        
        int[] DPstart = new int[N]; int[] DPend = new int[N];
        
        for (int S = 0; S < N; S++) {
            
            int counter = 0;
            for (int L = 0; L < N; L++) if (DP[S][L]) counter++;
            
            DPstart[S] = counter;
        }
        
        for (int L = 0; L < N; L++) {
            
            int counter = 0;
            for (int S = 0; S < N; S++) if (DP[S][L]) counter++;
            
            DPend[L] = counter;
        }
        
        long counter = 0; 
        for (int i = 0; i < N; i++) {
            for (int j = i + 1; j < N; j++) counter += DPend[i] * DPstart[j];
        }
        
        return counter;
    }
    

    
    /*private static long solve(String input) {
        
        char[] chars = input.toCharArray();
        
        
        
        
        
        
        
        
        
        
        
        
        
        int N = chars.length;
        long[] start = new long[N], end = new long[N];
        
        // Count # of palindromes of odd length
        for (int M = 0; M < N; M++) {
            for (int D = 0; M - D >= 0 && M + D < N && chars[M - D] == chars[M + D]; D++) {
                start[M - D]++; end[M + D]++;
            }
        }
        
        // Count # of palindromes of even length
        for (int M = 0; M < N + 1; M++) {
            for (int D = 0; M - D >= 0 && M + 1 + D < N && chars[M - D] == chars[M + 1 + D]; D++) {
                start[M - D]++; end[M + 1 + D]++;
            }
        }
        
        long counter = 0; 
        for (int i = 0; i < N; i++) {
            for (int j = i + 1; j < N; j++) counter += end[i] * start[j];
        }
        
        return counter;
    }*/

    public static void main(String[] args) {
        
        Scanner scanner = new Scanner(System.in);
        
        System.out.println(new PalindromePairs(scanner.next()).solve());
    }
}