#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const long long MAX = 1e18 + 1;
const double pi = 3.1415926535897932384626433832;
const long long M = 1e9 + 7;
long long bis(long long a[], long long l, long long r, long long x) {
  while (l <= r) {
    long long m = l + (r - l) / 2;
    if (a[m] == x) return m;
    if (a[m] < x)
      l = m + 1;
    else
      r = m - 1;
  }
  return -1;
}
long long gcd(long long x, long long y) {
  if (x == 0)
    return y;
  else
    return gcd(y % x, x);
}
long long lcm(long long x, long long y) { return (x * y) / gcd(x, y); }
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6) {
    if (n % i == 0 || n % (i + 2) == 0) {
      return false;
    }
  }
  return true;
}
bool pdrome(string s) {
  long long n = s.length();
  for (long long i = 0; i < n / 2; i++) {
    if (s[i] != s[n - i - 1]) {
      return false;
    }
  }
  return true;
}
string need(string s) {
  string a = s;
  reverse(a.begin(), a.end());
  a = s + '?' + a;
  long long n = a.size();
  long long pref[n];
  memset(pref, 0, sizeof(pref));
  for (long long i = 1; i < n + 1; i++) {
    long long len = pref[i - 1];
    while (len > 0 && a[len] != a[i]) len = pref[len - 1];
    if (a[i] == a[len]) len++;
    pref[i] = len;
  }
  return a.substr(0, pref[n - 1]);
}
long long mindig(long long x) {
  long long ans = 9;
  long long t = x;
  while (x) {
    ans = min(ans, x % 10);
    x /= 10;
  }
  x = t;
  return ans;
}
long long maxdig(long long x) {
  long long ans = 0;
  long long t = x;
  while (x) {
    ans = max(ans, x % 10);
    x /= 10;
  }
  x = t;
  return ans;
}
long long modpow(long long x, long long n, long long M) {
  if (n == 0) return 1 % M;
  long long u = modpow(x, n / 2, M);
  u = (u * u) % M;
  if (n % 2 == 1) u = (u * x) % M;
  return u;
}
long long sumdigits(long long a) {
  long long result = 0;
  while (a > 0) {
    result += a % 10;
    a /= 10;
  }
  return result;
}
long long digits(long long n) { return floor(log10(n) + 1); }
const long long Node = 1e5 + 7;
vector<long long> adj[Node];
bool vis[Node] = {false};
void addedge(long long u, long long v) {
  adj[u].push_back(v);
  adj[v].push_back(u);
}
void bfs(long long u) {
  vis[u] = true;
  list<long long> q;
  q.push_back(u);
  while (!q.empty()) {
    q.pop_front();
    for (auto x : adj[u]) {
      if (!vis[x]) {
        vis[x] = true;
        q.push_back(x);
      }
    }
  }
}
void dfs(long long u) {
  vis[u] = true;
  cout << u << " ";
  for (auto x : adj[u]) {
    if (!vis[x]) {
      dfs(x);
    }
  }
}
long long con1(string s) {
  long long res = 0;
  stringstream geek(s);
  geek >> res;
  return res;
}
string con2(long long num) {
  string stri = to_string(num);
  return stri;
}
long long countbits(long long k) {
  long long cnt = log2(k);
  return cnt;
}
bool check(string s, long long k1, long long j) {
  for (long long i = j; i < k1 / 2 + j; i++) {
    if (s[i] != s[i + k1 / 2]) return false;
  }
  return true;
}
void solve() {
  long long n;
  cin >> n;
  string s;
  cin >> s;
  long long ans = 2;
  if (n == 1) {
    cout << 1 << endl;
    return;
  }
  for (long long i = 0; i < 1; i++) {
    for (long long j = n; j >= i + 2; j--) {
      long long k = j - i;
      if (k % 2 == 0) {
        if (check(s, k, i) == true) {
          ans = max(ans, k);
        }
      }
    }
  }
  ans = n - ((ans) / 2 - 1);
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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

7
abcabca


Output

5


Input

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.

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const long long MAX = 1e18 + 1;
const double pi = 3.1415926535897932384626433832;
const long long M = 1e9 + 7;
long long bis(long long a[], long long l, long long r, long long x) {
  while (l <= r) {
    long long m = l + (r - l) / 2;
    if (a[m] == x) return m;
    if (a[m] < x)
      l = m + 1;
    else
      r = m - 1;
  }
  return -1;
}
long long gcd(long long x, long long y) {
  if (x == 0)
    return y;
  else
    return gcd(y % x, x);
}
long long lcm(long long x, long long y) { return (x * y) / gcd(x, y); }
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6) {
    if (n % i == 0 || n % (i + 2) == 0) {
      return false;
    }
  }
  return true;
}
bool pdrome(string s) {
  long long n = s.length();
  for (long long i = 0; i < n / 2; i++) {
    if (s[i] != s[n - i - 1]) {
      return false;
    }
  }
  return true;
}
string need(string s) {
  string a = s;
  reverse(a.begin(), a.end());
  a = s + '?' + a;
  long long n = a.size();
  long long pref[n];
  memset(pref, 0, sizeof(pref));
  for (long long i = 1; i < n + 1; i++) {
    long long len = pref[i - 1];
    while (len > 0 && a[len] != a[i]) len = pref[len - 1];
    if (a[i] == a[len]) len++;
    pref[i] = len;
  }
  return a.substr(0, pref[n - 1]);
}
long long mindig(long long x) {
  long long ans = 9;
  long long t = x;
  while (x) {
    ans = min(ans, x % 10);
    x /= 10;
  }
  x = t;
  return ans;
}
long long maxdig(long long x) {
  long long ans = 0;
  long long t = x;
  while (x) {
    ans = max(ans, x % 10);
    x /= 10;
  }
  x = t;
  return ans;
}
long long modpow(long long x, long long n, long long M) {
  if (n == 0) return 1 % M;
  long long u = modpow(x, n / 2, M);
  u = (u * u) % M;
  if (n % 2 == 1) u = (u * x) % M;
  return u;
}
long long sumdigits(long long a) {
  long long result = 0;
  while (a > 0) {
    result += a % 10;
    a /= 10;
  }
  return result;
}
long long digits(long long n) { return floor(log10(n) + 1); }
const long long Node = 1e5 + 7;
vector<long long> adj[Node];
bool vis[Node] = {false};
void addedge(long long u, long long v) {
  adj[u].push_back(v);
  adj[v].push_back(u);
}
void bfs(long long u) {
  vis[u] = true;
  list<long long> q;
  q.push_back(u);
  while (!q.empty()) {
    q.pop_front();
    for (auto x : adj[u]) {
      if (!vis[x]) {
        vis[x] = true;
        q.push_back(x);
      }
    }
  }
}
void dfs(long long u) {
  vis[u] = true;
  cout << u << " ";
  for (auto x : adj[u]) {
    if (!vis[x]) {
      dfs(x);
    }
  }
}
long long con1(string s) {
  long long res = 0;
  stringstream geek(s);
  geek >> res;
  return res;
}
string con2(long long num) {
  string stri = to_string(num);
  return stri;
}
long long countbits(long long k) {
  long long cnt = log2(k);
  return cnt;
}
bool check(string s, long long k1, long long j) {
  for (long long i = j; i < k1 / 2 + j; i++) {
    if (s[i] != s[i + k1 / 2]) return false;
  }
  return true;
}
void solve() {
  long long n;
  cin >> n;
  string s;
  cin >> s;
  long long ans = 2;
  if (n == 1) {
    cout << 1 << endl;
    return;
  }
  for (long long i = 0; i < 1; i++) {
    for (long long j = n; j >= i + 2; j--) {
      long long k = j - i;
      if (k % 2 == 0) {
        if (check(s, k, i) == true) {
          ans = max(ans, k);
        }
      }
    }
  }
  ans = n - ((ans) / 2 - 1);
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```