// C++ program to find palindromic substrings of a string
#include<bits/stdc++.h>
using namespace std;

// Returna total number of palindrome substring of
// length greater then equal to 2
int CountPS(char str[], int n)
{
	// creat empty 2-D matrix that counts all palindrome
	// substring. dp[i][j] stores counts of palindromic
	// substrings in st[i..j]
	int dp[n][n];
    printf("2\n");
	memset(dp, 0, sizeof(dp));

	// P[i][j] = true if substring str[i..j] is palindrome,
	// else false
	bool P[n][n];
	memset(P, false , sizeof(P));

	// palindrome of single lenght
	for (int i= 0; i< n; i++)
		P[i][i] = true;

	// palindrome of length 2
	for (int i=0; i<n-1; i++)
	{
		if (str[i] == str[i+1])
		{
			P[i][i+1] = true;
			dp[i][i+1] = 1 ;
		}
	}

	// Palindromes of length more then 2. This loop is similar
	// to Matrix Chain Multiplication. We start with a gap of
	// length 2 and fill DP table in a way that gap between
	// starting and ending indexes increases one by one by
	// outer loop.
	for (int gap=2 ; gap<n; gap++)
	{
		// Pick starting point for current gap
		for (int i=0; i<n-gap; i++)
		{
			// Set ending point
			int j = gap + i;

			// If current string is palindrome
			if (str[i] == str[j] && P[i+1][j-1] )
				P[i][j] = true;

			// Add current palindrome substring ( + 1)
			// and rest palinrome substring (dp[i][j-1] + dp[i+1][j])
			// remove common palinrome substrings (- dp[i+1][j-1])
			if (P[i][j] == true)
				dp[i][j] = dp[i][j-1] + dp[i+1][j] + 1 - dp[i+1][j-1];
			else
				dp[i][j] = dp[i][j-1] + dp[i+1][j] - dp[i+1][j-1];
		}
	}

	// return total palindromic substrings
	return dp[0][n-1];
}



// Driver program
int main()
{
	char str[] = "week";
	int n = strlen(str);
	cout << CountPS(str, n) << endl;
    
	return 0;
    
}