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: Accepted)

#include<iostream>
#include<string>
#include<vector>
#include<cmath>

using namespace std;

#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,s,n) for(long long i=s;i<n;i++)
#define all(a) (a).begin(),(a).end()

typedef long long ll;
typedef long double ld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
const double pi = acos(-1.0);
const ll mod = pow(10,9) + 7;
const ll INF = pow(2,31) - 1;

ll gcd(ll x,ll y)
{
    ll r;
    while((r = x % y) != 0)
    {
        x = y;
        y = r;
    }
    return y;
}

ll lcm(ll x,ll y)
{
    x /= gcd(x,y);
    y /= gcd(x,y);
    return (x*y);
}

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

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

	bool ok = true;

	//prime test
	ll i=x,ans;
	while(ok){
		REP(j,2,sqrt(i)+1){
			if(i % j == 0){
				ok = false;
				break;
			}
		}
		
		if(ok){
			ans = i;
			break;
		}
		else ok = true;
		
		i++;
	}

	cout << ans << endl;

	return 0;
}