/*

  答案为n-fail[n]

  我们可以分n-fail[n]>=n/2和n-fail[n]<n/2来讨论，画画图就可以得出结论。

*/

#include<cstdio>

#include<iostream>

#define N 1000010

using namespace std;

int fail[N],n;

char s[N];

int main(){

    scanf("%d%s",&n,s+);

    fail[]=;

    for(int i=;i<=n;i++){

        int p=fail[i-];

        while(p&&s[p+]!=s[i]) p=fail[p];

        if(s[p+]==s[i]) fail[i]=p+;

        else fail[i]=;

    }

    printf("%d",n-fail[n]);

    return ;

}