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>

#define forn(i,a,b) for(int i=(a); i<(b); i++)
#define all(x) begin(x),end(x)
#define ll long long
#define vi vector<int>

using namespace std;

const int NMAX=1000001;
vi isPrime(NMAX,1);

void fillP(int n){
        isPrime[0]=0,isPrime[1]=0;
        for(int i=2; i<=n; i++)
                if(isPrime[i])
                        for(int j=2*i; j<=n; j+=i)
                                isPrime[j]=false;
}



int main(){
        ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
        int a,ans;
        cin>>a;
        fillP(NMAX);
        ans=a;
        while(!isPrime[ans]){
                ans++;
        }
        cout<<ans;
        return 0;
}