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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdbool.h>
#include <math.h>

#define MAX 200000 //ベルトラン・チェビシェフの定理より,100000以降の初めの素数は必ず200000以下にあるから.

bool solve(int X);
int X;

int main(int argc, char const *argv[])
{
	scanf("%d", &X);
	for (int i = X; i < MAX; ++i)
	{
		if (solve(i))
		{
			printf("%d\n", i);
			break;
		}
	}

	return 0;
}



bool solve(int X){
	if (X%2 == 0)
	{
		return false;
	}

	else{
		int x;
		x = (int)sqrt(X);
		for (int i = 3; i < x+1; i++)
		{
			if (X%i == 0)
			{
				return false;
			}
		}

		return true;
	}

}