Problem:
Score: 300 points

Problem Statement

Find the minimum prime number greater than or equal to X.

Notes

A prime number is an integer greater than 1 that cannot be evenly divided by any positive integer except 1 and itself.

For example, 2, 3, and 5 are prime numbers, while 4 and 6 are not.

Constraints

2 \le X \le 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X

Output

Print the minimum prime number greater than or equal to X.

Sample Input 1

20

Sample Output 1

23

The minimum prime number greater than or equal to 20 is 23.

Sample Input 2

2

Sample Output 2

2

X itself can be a prime number.

Sample Input 3

99992

Sample Output 3

100003

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

#include <bits/stdc++.h>

using namespace std;

inline int ctoi(char c) { if(c < '0' || '9' < c) throw invalid_argument("ctoi error"); return c - '0'; }
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define REP(i, k, n) for(int i = (int)(k); i < (int)(n); i++)
#define all(x) (x).begin(), (x).end()

template <typename T>
inline T gcd(T x, T y){
	if (x <= 0 || y <= 0) throw invalid_argument("gcd error: x <= 0 or y <= 0");
	
	if(x < y) swap(x, y);
	T r = x % y;

	while(r != 0){
		x = y;
		y = r;
		r = x % y;
	}

	return y;
}
template <typename T>
inline T lcm(T x, T y){
	if (x <= 0 || y <= 0) throw invalid_argument("lcm error: x <= 0 or y <= 0");

	return x * y / gcd(x, y);
}

int main(){
	int x;
	cin >> x;

	if(x == 2){
		cout << 2;
		return 0;
	}

	int flg = true;

	while(flg){
		int n = (int)sqrt(x);

		int i;
		for(i = 2; i <= n; i++){
			if(x % i == 0){
				flg = false;
				break;
			}
		}

		if(i == n + 1){
			break;
		}
		else{
			x++;
		}
	}

	cout << x;
	return 0;
}