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.*;
import java.lang.*;

public class D1 {
	public static char[] string;

	public static void main(String[] args) {
		FastReader in = new FastReader();
		PrintWriter out = new PrintWriter(System.out);
		string = in.next().toCharArray();
		int n = string.length;
		int start[] = new int[string.length];
		int end[] = new int[string.length];
		for(int i=0;i<string.length;i++) {
			for(int j=i;j<string.length;j++) {
				if(isPalindrome(i,j)) {
					start[i]++;
					end[j]++;
				}
			}
		}
		long ans = 0;
		for(int i=n-2;i>=0;i--) start[i]+=start[i+1];
		for(int i=0;i<n-1;i++) ans+= end[i]*start[i+1];
		out.println(ans);
		out.close();
	}
	public static boolean isPalindrome(int left,int right) {
		while(left<=right) {
			if(string[left]==string[right]) {
				left++;
				right--;
			}
			else return false;
		}
		return true;
	}

	static final Random random=new Random();
	//	static void ruffleSort(Pair[] a) {
	//		int n=a.length;//shuffle, then sort 
	//		for (int i=0; i<n; i++) {
	//			int oi=random.nextInt(n);
	//			Pair temp=a[oi];
	//			a[oi]=a[i]; a[i]=temp;
	//		}
	//		Arrays.sort(a);
	//	}
	static void ruffleSort(int[] a) {
		int n=a.length;//shuffle, then sort 
		for (int i=0; i<n; i++) {
			int oi=random.nextInt(n), temp=a[oi];
			a[oi]=a[i]; a[i]=temp;
		}
		Arrays.sort(a);
	}
	static void ruffleSort(char[] a) {
		int n=a.length;//shuffle, then sort 
		for (int i=0; i<n; i++) {
			int oi=random.nextInt(n);
			char temp=a[oi];
			a[oi]=a[i]; a[i]=temp;
		}
		Arrays.sort(a);
	}
	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; 
		} 

		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
	}
}

//class Pair implements Comparable<Pair>{
//    int a;
//    int b;
//    public Pair(int a, int b) {
//        this.a = a;
//        this.b = b;
//    }
//    public int compareTo(Pair o) {
//        if(this.a==o.a)
//            return this.b - o.b;
//        return this.a - o.a;
//    }   
//}