/*

Palindrome (IOI 2000)
Copyright (C) 2000 Paolo Boldi and Sebastiano Vigna

This program is free software; you can redistribute it and/or modify
it under the terms of the GNU General Public License as published by
the Free Software Foundation; either version 2 of the License, or
(at your option) any later version.

This program is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
GNU General Public License for more details.

You should have received a copy of the GNU General Public License
along with this program; if not, write to the Free Software
Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA  02111-1307  USA



Il programma si basa sull'idea che se f(w) e` il minimo numero di caratteri che
rende w palindroma, allora f(awa)=f(w) e f(awb) = min{f(wb), f(aw)}+1. La
dimostrazione e` data a parte. 

Dato che la matrice necessaria per calcolare per programmazione dinamica f e`
5000x5000, ma d'altra parte per calcolare la riga i ci servono solo le righe i-1 
e i-2, utilizziamo sempre le stesse tre righe da 5000 elementi ciclicamente.

*/


#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <limits.h>

#define min(a,b) ((a)<(b)?(a):(b))

#define MAXLUN 5000
#define MINLUN 3

char parola[MAXLUN];
int ott[3][MAXLUN]; /* Le tre righe usate ciclicamente per la programmazione dinamica. */
FILE *F;

int main(int argc, char *argv[]) {
	int k, i, N;
	FILE *F;

	F = stdin;
	fscanf(F, "%d\n", &N);
	for(i=0; i<N; i++) fscanf(F, "%c", &parola[i]);

	for(k=2; k<=N; k++)
		for(i=0; i<=N-k; i++) {
			if (parola[i] == parola[i+k-1]) ott[(k+2)%3][i] = ott[k%3][i+1];
			else ott[(k+2)%3][i] = 1+min(ott[(k+1)%3][i], ott[(k+1)%3][i+1]);
		}	

	F = stdout;

	fprintf(F, "%d\n", ott[(N+2)%3][0]);

	return 0;
}