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.*;
public class test
{
public static void main(String[] args)
{

Scanner sc=new Scanner(System.in);
String s=sc.next();
int n=s.length();
boolean p[][]=new boolean[n][n];
for(int i=0;i<n;i++)
p[i][i]=true;
for(int i=0;i<n-1;i++)
{
p[i][i+1]=(s.charAt(i)==s.charAt(i+1));
}
for(int gap=2;gap<n;gap++)
{
for(int j=0;j<n-gap;j++)
{
int i=j;
int k=i+gap;
p[i][k]=(p[i+1][k-1]&&(s.charAt(i)==s.charAt(k)));
}}
int stat[]=new int[n];
for(int i=0;i<n;i++)
{
int count=0;
for(int j=i;j<n;j++)
{
if(p[i][j])
count++;
}
stat[i]=count;}
int dp[]=new int[n];
dp[n-1]=1;
for(int i=n-2;i>=0;i--)
{
dp[i]=dp[i+1]+stat[i];}
long sum=0;
for(int i=0;i<n;i++)
{
for(int j=i;j<n;j++)
{
if(p[i][j])
{
if(j+1<n)
sum=sum+(long)dp[j+1];
else
sum=sum;}}}
System.out.println(sum);}}