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.*;
import java.util.StringTokenizer;

public class main159D {
    public static PrintWriter out = new PrintWriter(System.out);
    public static FastScanner enter = new FastScanner(System.in);

    public static void main(String[] args) throws IOException {
        solve();
        out.close();
    }

    static int dp[][];//indikator - [l, r]- палиндром
    static int dpl[]; // kol_vo pal begin with [l
    static int dpr[]; // kol_vo pal end with r]
    private static void solve() throws IOException {
        String s=enter.next();
        dp=new int[s.length()][s.length()];
        dpr=new int[s.length()];
        dpl=new int[s.length()];
        for (int i = 1; i <=s.length() ; i++) {
            for (int j = 0; j+i-1 <s.length() ; j++) {
                int left=j;
                int right=j+i-1;
                if(i==1) dp[left][right]=1;
                else dp[left][right]=(s.charAt(left)==s.charAt(right) && (dp[left+1][right-1]==1 || left+1>right-1))?1:0;
            }
        }
        for (int i = 0; i <s.length() ; i++) {
            for (int j = i; j <s.length() ; j++) {
                dpl[i]+=dp[i][j];
            }
        }
        for (int i = 0; i <s.length() ; i++) {
            for (int j = i; j >=0 ; j--) {
                dpr[i]+=dp[j][i];
            }
        }
        long ans=0;
        for (int i = 0; i <s.length() ; i++) {
            long summ=0;
            for (int j = 0; j <i ; j++) {
                summ+=dpr[j];
            }
            summ*=(long)dpl[i];
            ans+=summ;
        }
        out.println(ans);
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer stok;

        FastScanner(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

        String next() throws IOException {
            while (stok == null || !stok.hasMoreTokens()) {
                String s = br.readLine();
                if (s == null) {
                    return null;
                }
                stok = new StringTokenizer(s);
            }
            return stok.nextToken();
        }

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

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

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

        char nextChar() throws IOException {
            return (char) (br.read());
        }

        String nextLine() throws IOException {
            return br.readLine();
        }
    }
}