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 <bits/stdc++.h>
/*#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>*/
 
#define pll              pair <long long,long long>
#define pb               push_back
#define ll               long long
#define ff               first
#define ss               second
#define D(x)             cerr<<#x " = "<<(x)<<endl
#define vll              vector <ll>
#define IOS              ios_base::sync_with_stdio(0); cin.tie(NULL);
#define endl             '\n'   
using namespace std;
ll p[100005];
void seive()
{
	memset(p,0,sizeof(p));
	p[1]=1;
	for(ll i=2;i*i<100005;i++)
	{
		if(p[i])continue;
		for(ll j=i*i;j<100005;j+=i)
		{
			p[j]=1;
		}
	}
	return ;
}

int main()
{
    IOS
 	// freopen("input.txt","r",stdin);
 	// freopen("output.txt","w",stdout);
 	// freopen("error.txt","w",stderr);
 	seive();
   	ll n;
   	cin>>n;
   	for(ll i=n;i<100005;i++)
   	{
   		if(i>=n && !p[i])
   		{
   			cout<<i<<endl;
   			return 0;
   		}
   	}
    return 0;
}