Problem:
Score : 200 points

Problem Statement

Snuke is interested in strings that satisfy the following conditions:

The length of the string is at least N.

The first N characters equal to the string s.

The last N characters equal to the string t.

Find the length of the shortest string that satisfies the conditions.

Constraints

1≤N≤100

The lengths of s and t are both N.

s and t consist of lowercase English letters.

Input

The input is given from Standard Input in the following format:

N
s
t

Output

Print the length of the shortest string that satisfies the conditions.

Sample Input 1

3
abc
cde

Sample Output 1

5

The shortest string is abcde.

Sample Input 2

1
a
z

Sample Output 2

2

The shortest string is az.

Sample Input 3

4
expr
expr

Sample Output 3

4

The shortest string is expr.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
using ll =long long;
using db = double;
using vll = vector<long long>;

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
#define che(a,string) cout<<"//"<<string <<"=="<<(a)<<"//" <<endl;
#define SORT(a) sort((a).begin(),(a).end())
#define rSORT(a) reverse((a).begin(),(a).end())
#define For(i, a, b)    for(int i = (a) ; i < (b) ; ++i)
#define rep(i, n)       For(i, 0, n)
#define leftunique(a) {SORT(a);(a).erase(unique((a).begin(),(a).end()),(a).end());}
//leftuniqueは配列から同じ要素のものを取り除く 例 4 6 2 1 3 4 1 2 なら 1 2 3 4 6 になる
#define debug(x)  cout << #x << " = " << (x) << endl;
bool IsInt(double a){int b = a/1;if(a==b){return true;}else {return false;}}
bool coY() {cout <<"Yes"<<endl;}
bool coN(){cout <<"No"<<endl;}
const int mod = 1e9 + 7;
const ll INF = 1LL << 60;
// a^n mod を計算する
long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
//グラフ用
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};	
//ここから書き始める

ll combination(ll n, ll i){
	ll count = 0;
	ll multi = 1;
	ll div =1;
	int limit = i;
	while(count < limit){
		multi *= n;
		n--;
		multi/= count+1;
		count ++ ;
	}
	return multi / div;
}

int main()
{
	cout <<setprecision(10);
	ll n ; 
	string s , t;
	cin >> n >> s >> t;
	int additional = n;

	int j = 0 ;
	for(int i = 0 ; i < n ; i ++){
		if(s[i] == t[j]){
			additional --;
			j++;
		}else{
			additional = n;
			j = 0;
		}
	}
	cout << n + additional << endl;
}