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 CF;
// 
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class A {
	static boolean [][] cmp, memo;
	static char [] c;
	static boolean dp(int s, int t){
		if(s >= t)
			return true;
		if(cmp[s][t])
			return memo[s][t];
		boolean palin = (c[s] == c[t]) & dp(s+1, t-1);
		cmp[s][t] = true;
		return memo[s][t] = palin;
	}
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		c = sc.next().toCharArray();
		int n = c.length;
		cmp = new boolean[n+1][n+1];
		memo = new boolean[n+1][n+1];
		long ans = 0;
		long [] res = new long[n];
		res[0] = 1;
		for (int j = 1; j < n; ++j) {
			int cur = 0;
			for (int i = 0; i <= j; ++i) {
				if(dp(i, j)){
					++cur;
					if(i > 0)
						ans += res[i-1];
				}
			}
			res[j] = res[j-1] + cur;
		}
		out.println(ans);
		out.flush();
		out.close();
	}


	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(FileReader fileReader) {
			br = new BufferedReader(fileReader);
		}

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

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

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

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

		public boolean ready() throws IOException {
			return br.ready();
		}
	}
}