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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

//update 2019/12/13

public class Main{

  static Scanner sc = new Scanner(System.in);

  public static void main(String[] args) {

    int x = nextInt();

    int ans = 0;

  if(x==2){
    ans = 2;
  }
  else{
    while(true){
      int flag = 0;
      for(int i=2;i<x;i++){
        if(x%i==0){
          flag = 1;
          break;
        }
      }
      if(flag==0){
        break;
      }
      x++;
    }
  }

  System.out.println(x);


  }

  static String next(){
    return sc.next();
  }

  static int nextInt(){
    return Integer.parseInt(sc.next());
  }

  static long nextLong(){
    return Long.parseLong(sc.next());
  }

  static double nextDouble(){
    return Double.parseDouble(sc.next());
  }

  static String[] splitString(String s){
    return s.split("");
  }

  static char[] splitChar(String s){
    return s.toCharArray();
  }

  static int charToInt(char aaa){
    return aaa - 48;
  }

  public static int maxInt(int[] a,int lng){
    int max=a[0];
    for(int i=1;i<lng;i++){
      max = Math.max(max,a[i]);
    }
    return max;
  }

  public static int minInt(int[] a,int lng){
    int min=a[0];
    for(int i=1;i<lng;i++){
      min = Math.min(min,a[i]);
    }
    return min;
  }

  public static long maxLong(long[] a,int lng){
    long max=a[0];
    for(int i=1;i<lng;i++){
      max = Math.max(max,a[i]);
    }
    return max;
  }

  public static long minLong(long[] a,int lng){
    long min=a[0];
    for(int i=1;i<lng;i++){
      min = Math.min(min,a[i]);
    }
    return min;
  }

}