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 (GCC 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define rep(i,n) for(i=0;i<n;i++)
#define PI 3.14159265358979323846264338327950L
// xが素数か判定する(素数なら1, 非素数は0)
int prime( int x )
{
    int n;

    if( x == 2 ) return 1;
    if( x < 2 || x % 2 == 0 ) return 0;
    for( n = 3; n <= sqrt((double)x); n += 2 )
        if( x % n == 0 ) return 0;
    return 1;
}
int main(){
    int i,x;
    scanf("%d",&x);
    for(i=0;i<x;i++){
        if(prime(x+i)) break; 
    }
    printf("%d",x+i);
    return 0;
}