Problem:
Score: 200 points

Problem Statement

We have a string S consisting of lowercase English letters.

If the length of S is at most K, print S without change.

If the length of S exceeds K, extract the first K characters in S, append ... to the end of them, and print the result.

Constraints

K is an integer between 1 and 100 (inclusive).

S is a string consisting of lowercase English letters.

The length of S is between 1 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

K
S

Output

Print a string as stated in Problem Statement.

Sample Input 1

7
nikoandsolstice

Sample Output 1

nikoand...

nikoandsolstice has a length of 15, which exceeds K=7.

We should extract the first 7 characters in this string, append ... to the end of them, and print the result nikoand....

Sample Input 2

40
ferelibenterhominesidquodvoluntcredunt

Sample Output 2

ferelibenterhominesidquodvoluntcredunt

The famous quote from Gaius Julius Caesar.

Solution in C++ (Clang 10.0.0) (judge verdict: Accepted)

#include<iostream>
#include<string>
#include<vector>
#include<utility>
#include<algorithm>
#include<map>
#include<cstdlib>
#include<cmath>
#include<numeric>
#include<iomanip>
#include<functional>
#include<cstdlib>

#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
using ll =long long;
const int mod = 1000000007;
map<string, ll>a;
class mint {
	long long x;
public:
	mint(long long x = 0) : x((x% mod + mod) % mod) {}
	mint operator-() const {
		return mint(-x);
	}
	mint& operator+=(const mint & a) {
		if ((x += a.x) >= mod) x -= mod;
		return *this;
	}
	mint & operator-=(const mint & a) {
		if ((x += mod - a.x) >= mod) x -= mod;
		return *this;
	}
	mint & operator*=(const  mint & a) {
		(x *= a.x) %= mod;
		return *this;
	}
	mint operator+(const mint & a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint & a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint & a) const {
		mint res(*this);
		return res *= a;
	}
	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}
	// for prime mod
	mint inv() const {
		return pow(mod - 2);
	}
	mint& operator/=(const mint & a) {
		return (*this) *= a.inv();
	}
	mint operator/(const mint & a) const {
		mint res(*this);
		return res /= a;
	}

	friend ostream& operator<<(ostream & os, const mint & m) {
		os << m.x;
		return os;
	}
};
int ctoi(char c) {
	if (c >= '0' && c <= '9') {
		return c - '0';
	}
	return 0;
}
//組み合わせ
void recursive_comb(vector<int> indexes, int s, int rest, std::function<void(vector<int>)> f) {
	if (rest == 0) {
		f(indexes);
	}
	else {
		if (s < 0) return;
		recursive_comb(indexes, s - 1, rest, f);
		indexes[rest - 1] = s;
		recursive_comb(indexes, s - 1, rest - 1, f);
	}
}

// nCkの組み合わせに対して処理を実行する
void foreach_comb(int n, int k, std::function<void(vector<int>)> f) {
	vector<int>indexes(k);
	recursive_comb(indexes, n - 1, k, f);
}
string num[10] = { "pon","pon","hon","bon","hon","hon","pon","hon","pon","hon" };

int main() {
	string S;
	ll K;
	cin >> K >> S;
	if (S.size() > K) {
		cout << S.substr(0,K) << "...";

	}
	else {
		cout << S;
	}
}