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:
print(function(s) {
	var i, j,
		l = s.length,
		isP = [];

	for (i = 0; i < l; i++)
		isP[i] = [];

	for (i = 0; i < l; i++)
		for (j = 0; j < l; j++)
			isP[i][j] = 0;

	isP[0][0] = 1;


	for (i = l; i >= 0; i--)
		for (j = i; j < l; j++)
			if (s[i] === s[j])
				isP[i][j] = j - i > 1 ? isP[i + 1][j - 1] : 1;
			else
				isP[i][j] = 0;

	var sum = [1];
	for (j = 1; j < l; j++) {
		sum[j] = sum[j - 1];
		for (i = 0; i <= j; i++)
			sum[j] += isP[i][j];
	}

	var ans = 0;
	for (j = 1; j < l; j++)
		for (i = 1; i <= j; i++)
			if (isP[i][j] === 1)
				ans += sum[i - 1];

	return ans;
}(readline()));