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:
//package vkcup1.q2;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;

public class D {
	InputStream is;
	PrintWriter out;
	String INPUT = "";
	
	void solve()
	{
		char[] s = ns(2000);
		int[] pa = palindrome(s);
		long ct = 0;
		for(int i = 0;i < pa.length;i++){
			for(int j = i+1;j < pa.length;j++){
				if(pa[i] == 0 || pa[j] == 0)continue;
				int d = ((j-(j%2==0?1:2))-(i+(i%2==0?1:2)))/2;
				int x = Math.min(d, (pa[i]-(i%2==0?1:2))/2);
				int y = Math.min(d, (pa[j]-(j%2==0?1:2))/2);
				if(d >= 0){
					long lct = 0;
					if(x+y<=d){
						lct = (long)(y+1)*(x+1);
					}else{
						lct += (long)(d+1)*(d+2)/2;
						lct -= (long)(d-y)*(d-y+1)/2;
						lct -= (long)(d-x)*(d-x+1)/2;
					}
					ct += lct;
				}
			}
		}
		out.println(ct);
	}
	
	public static int[] palindrome(char[] str)
	{
		int n = str.length;
		int[] r = new int[2*n];
		int k = 0;
		for(int i = 0, j = 0;i < 2*n;i += k, j = Math.max(j-k, 0)){
			// 普通の回文検査
			while(i-j >= 0 && i+j+1 < 2*n && str[(i-j)/2] == str[(i+j+1)/2])j++;
			r[i] = j;
			
			// 定理に基づいて回文検査を端折る
			for(k = 1;i-k >= 0 && r[i]-k >= 0 && r[i-k] != r[i]-k;k++){
				r[i+k] = Math.min(r[i-k], r[i]-k);
			}
		}
		return r;
	}
	
	void run() throws Exception
	{
		is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);
		
		long s = System.currentTimeMillis();
		solve();
		out.flush();
		tr(System.currentTimeMillis()-s+"ms");
	}
	
	public static void main(String[] args) throws Exception
	{
		new D().run();
	}
	
	public int ni()
	{
		try {
			int num = 0;
			boolean minus = false;
			while((num = is.read()) != -1 && !((num >= '0' && num <= '9') || num == '-'));
			if(num == '-'){
				num = 0;
				minus = true;
			}else{
				num -= '0';
			}
			
			while(true){
				int b = is.read();
				if(b >= '0' && b <= '9'){
					num = num * 10 + (b - '0');
				}else{
					return minus ? -num : num;
				}
			}
		} catch (IOException e) {
		}
		return -1;
	}
	
	public long nl()
	{
		try {
			long num = 0;
			boolean minus = false;
			while((num = is.read()) != -1 && !((num >= '0' && num <= '9') || num == '-'));
			if(num == '-'){
				num = 0;
				minus = true;
			}else{
				num -= '0';
			}
			
			while(true){
				int b = is.read();
				if(b >= '0' && b <= '9'){
					num = num * 10 + (b - '0');
				}else{
					return minus ? -num : num;
				}
			}
		} catch (IOException e) {
		}
		return -1;
	}
	
	public String ns()
	{
		try{
			int b = 0;
			StringBuilder sb = new StringBuilder();
			while((b = is.read()) != -1 && (b == '\r' || b == '\n' || b == ' '));
			if(b == -1)return "";
			sb.append((char)b);
			while(true){
				b = is.read();
				if(b == -1)return sb.toString();
				if(b == '\r' || b == '\n' || b == ' ')return sb.toString();
				sb.append((char)b);
			}
		} catch (IOException e) {
		}
		return "";
	}
	
	public char[] ns(int n)
	{
		char[] buf = new char[n];
		try{
			int b = 0, p = 0;
			while((b = is.read()) != -1 && (b == ' ' || b == '\r' || b == '\n'));
			if(b == -1)return null;
			buf[p++] = (char)b;
			while(p < n){
				b = is.read();
				if(b == -1 || b == ' ' || b == '\r' || b == '\n')break;
				buf[p++] = (char)b;
			}
			return Arrays.copyOf(buf, p);
		} catch (IOException e) {
		}
		return null;
	}
	
	
	double nd() { return Double.parseDouble(ns()); }
	boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); }
}