string substr (size_t pos = 0, size_t len = npos) const;

這個函式的作用是取字串string中位置為pos開始的len個字元，返回一個字串。

如果pos開始後len個字元長度超出了字串的長度，則取到最後一個字元。

下面給出示範：

#include <iostream>
#include <string>
using namespace std;
int main ()
{
  string str="We think in generalities, but we live in details.";
                                           // (quoting Alfred N. Whitehead)

  string str2 = str.substr (3,5);     // "think"

  size_t pos = str.find("live");      // position of "live" in str

  string str3 = str.substr (pos);     // get from "live" to the end

  cout << str2 << ' ' << str3 << '\n';

  return 0;
}
 

最後輸出為

think live in details.

You are given a string s consisting of n lowercase Latin letters. You have to type this string using your keyboard.

Initially, you have an empty string. Until you type the whole string, you may perform the following operation:

• add a character to the end of the string.

Besides, at most once you may perform one additional operation: copy the string and append it to itself.

For example, if you have to type string abcabca, you can type it in 7 operations if you type all the characters one by one. However, you can type it in 5 operations if you type the string abc first and then copy it and type the last character.

If you have to type string aaaaaaaaa, the best option is to type 4 characters one by one, then copy the string, and then type the remaining character.

Print the minimum number of operations you need to type the given string.

Input

The first line of the input containing only one integer number n (1 ≤ n ≤ 100) — the length of the string you have to type. The second line containing the string s consisting of n lowercase Latin letters.

Output

Print one integer number — the minimum number of operations you need to type the given string.

Examples

input

Copy

7
abcabca

output

5

input

Copy

8
abcdefgh

output

8

Note

The first test described in the problem statement.

In the second test you can only type all the characters one by one.

就可以用substr減輕程式碼量啦~

#include <iostream>
#include <string>
using namespace std;
int main ()
{
    int n;
    string s;
    cin>>n;
    cin>>s;
    int num = 1;
    for(int i = 1;i < n; ++i)
    {
        if(s.substr(0,i) == s.substr(i,i))
            num = i;
    }
    num = n - num + 1;
    cout<<num<<endl;
  return 0;
}