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;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Solver {

    StringTokenizer st;
    BufferedReader in;
    PrintWriter out;

    public static void main(String[] args) throws NumberFormatException, IOException {
        Solver solver = new Solver();
        solver.open();
        solver.solve();
        solver.close();
    }

    public void open() throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
    }

    public String nextToken() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }

    public int nextInt() throws NumberFormatException, IOException {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws NumberFormatException, IOException {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws NumberFormatException, IOException {
        return Double.parseDouble(nextToken());
    }

    public void solve() throws NumberFormatException, IOException {
        String buf = nextToken();
        char[] s = buf.toCharArray();
        int n = buf.length();

        int[] d1 = new int[n];
        int[] d2 = new int[n];

        int l = 0, r = -1;
        for (int i = 0; i < n; ++i) {
            int k = (i > r ? 0 : Math.min(d1[l + r - i], r - i)) + 1;
            while (i + k < n && i - k >= 0 && s[i + k] == s[i - k])
                ++k;
            d1[i] = --k;
            if (i + k > r) {
                l = i - k;
                r = i + k;
            }
        }

        l = 0;
        r = -1;
        for (int i = 0; i < n; ++i) {
            int k = (i > r ? 0 : Math.min(d2[l + r - i + 1], r - i + 1)) + 1;
            while (i + k - 1 < n && i - k >= 0 && s[i + k - 1] == s[i - k])
                ++k;
            d2[i] = --k;
            if (i + k - 1 > r) {
                l = i - k;
                r = i + k - 1;
            }
        }
        
        int[] startHere = new int[n];
        int[] finishHere = new int[n];
        
        for (int i=0;i<n;i++){
            int last = d1[i];
            for (int j=i-1;last>0;j--,last--){
                startHere[j]++;
            }
            last = d2[i];
            for (int j=i-1;last>0;j--,last--){
                startHere[j]++;
            }
            startHere[i]++;
            
            last = d1[i];
            for (int j=i+1;last>0;j++,last--){
                finishHere[j]++;
            }
            last = d2[i];
            for (int j=i;last>0;j++,last--){
                finishHere[j]++;
            }
            finishHere[i]++;
        }
        
        long[] allAfter = new long[n];
        
        long sum = 0;
        for (int i=n-1;i>=0;i--){
            allAfter[i] = sum;
            sum += startHere[i];
        }
        
        long result =0;
        
        for (int i=0;i<n;i++){
            result += finishHere[i]*allAfter[i];
        }
        
        out.println(result);
    }

    public void close() {
        out.flush();
        out.close();
    }

}