埃及筛法找到所有范围内的素数，然后直接lowerbound即可，

#define ll int
#define MAX 20005
#define vec vector<ll>

int main() {
	bool isP[MAX]; vec primes;
	fill(isP, isP + MAX, 1);
	for (int i = 2; i < MAX; i++) {
		if (isP[i])
			for (int j = i * 2; j < MAX; j += i)
				isP[j] = 0;
	}
	for (int i = 2; i < MAX; i++)if (isP[i])primes.push_back(i);
	ll n;
	while (cin >> n) {
		ll id = lower_bound(primes.begin(), primes.end(), n) - primes.begin();
		cout << primes[id] << endl;
	}
}