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>
using namespace std;
using ll = long long;
//const ll MOD = 1e9 + 7;
//const ll MOD = 998244353;
//const ll MOD = ;
ll mod(ll A, ll M) {return (A % M + M) % M;}
const ll INF = 1LL << 60;
template<class T> bool chmin(T& a, T b) {if (a > b) {a = b; return true;} return false;}
template<class T> bool chmax(T& a, T b) {if (a < b) {a = b; return true;} return false;}
ll divceil(ll A, ll B) {return (A + (B - 1)) / B;}
ll myctoi(char C) {return C - '0';}
char myitoc(ll N) {return '0' + N;}
#define FINALANS(A) {cout << (A) << '\n'; exit(0);}

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

  vector<int> P;
  vector<bool> Pb(111111, false);
  Pb.at(2) = true;
  for (int p = 3; p < 111111; p += 2)
  {
    bool ok = true;
    for (int i = 0; i < P.size(); i++)
    {
      int PP = P.at(i);
      if (PP * PP > p) break;
      if (p % PP == 0)
      {
        ok = false;
        break;
      }
    }
    if (ok)
    {
      P.push_back(p);
      Pb.at(p) = true;
    }
  }
  
  for (ll i = X; i < Pb.size(); i++)
  {
    if (Pb.at(i))
      FINALANS(i);
  }
}