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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
 #include <limits.h>
int main(void)
{
    int X;
    scanf("%d",&X);
    printf("%d",NextPrime(X,INT_MAX));
    return 0;
}
/*素数を判定する関数
 * 素数であれば1を返し、素数でなければ0を返す
 */
int prime(int n) {
	int i;

	if (n <= 1)/*1以下の場合は0を返す*/
	{
		return 0;/*素数ではない*/
	}
	/*nまでの数値を繰り返し、余りがあれば0を返す*/
	for (i = 2; i < n; i++) {
		if (n % i == 0) {
			return 0;/*素数ではない*/
		}
	}

	return 1;/*素数である*/
}
int NextPrime(int value,int max)
{
        int i;
    for( i=value;i<max;++i ) {
    if( prime(i)==1 ) {
      return i;
    }
  }
  return -1;

}