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.awt.*;
import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class Abc {
    static char[] s;
    static int dp[][];
    public static void main(String[] args) throws IOException {
        FastReader sc = new FastReader();
        s=sc.next().toCharArray();
        dp=new int[s.length][s.length+1];
        for (int arr[]:dp) Arrays.fill(arr,-1);
        int sum[]=new int[s.length+1];
        for (int l=1;l<=s.length;l++){
            for (int i=0;i<=s.length-l;i++){
                if (dp(i,l)==1) {
                    sum[i + l-1] +=1;
                }
            }
        }
        for (int i=1;i<s.length;i++){
            sum[i]+=sum[i-1];
        }
        long ans=0;
        for (int i=1;i<s.length;i++){
            for (int j=s.length-i;j>=1;j--){
                if (dp(i,j)==1){
                    ans+=sum[i-1];
                }
            }
        }
        System.out.println(ans);

    }
    static int dp(int i,int l){
        if (l==1||l==0)return 1;
        if (dp[i][l]!=-1)return dp[i][l];
        return dp[i][l]=dp(i+1,l-2)==1 && s[i]==s[i+l-1]?1:0;
    }
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}