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)

#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>
#define FASTIO ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define rep(i, l, r) for (ll i = (l); i < (r); ++i)
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define pii pair<int, int>
#define pll pair<ll, ll>
#define pli pair<ll, int> 
#define pdd pair<ld, ld> 
#define pb push_back
#define reset(x,v) memset(x,v,sizeof(x))
#define mod 1000000007
using namespace std;
set<int> pr;
void SieveOfEratosthenes(int n) 
{ 
    // Create a boolean array "prime[0..n]" and initialize 
    // all entries it as true. A value in prime[i] will 
    // finally be false if i is Not a prime, else true. 
    bool prime[n+1]; 
    memset(prime, true, sizeof(prime)); 
  
    for (int p=2; p*p<=n; p++) 
    { 
        // If prime[p] is not changed, then it is a prime 
        if (prime[p] == true) 
        { 
            // Update all multiples of p greater than or  
            // equal to the square of it 
            // numbers which are multiple of p and are 
            // less than p^2 are already been marked.  
            for (int i=p*p; i<=n; i += p) 
                prime[i] = false; 
        } 
    } 
  
    // Print all prime numbers 
    for (int p=2; p<=n; p++) 
       if (prime[p]) 
          pr.insert(p); 
} 

int main(){
    FASTIO
    int n;
    cin >> n;
    SieveOfEratosthenes(1000000);
    cout << *(pr.lower_bound(n));
     return 0;
}