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: Wrong Answer)

#include<iostream>
using namespace std;
int modulo(int a,int b,int c){
    long long x=1,y=a;
    while(b > 0){
        if(b%2 == 1){
            x=(x*y)%c;
        }
        y = (y*y)%c; 
        b /= 2;
    }
    return x%c;
}
long long mulmod(long long a,long long b,long long c){
    long long x = 0,y=a%c;
    while(b > 0){
        if(b%2 == 1){
            x = (x+y)%c;
        }
        y = (y*2)%c;
        b /= 2;
    }
    return x%c;
}
bool Miller(long long p,int iteration){
    if(p<2){
        return false;
    }
    if(p!=2 && p%2==0){
        return false;
    }
    long long s=p-1;
    while(s%2==0){
        s/=2;
    }
    for(int i=0;i<iteration;i++){
        long long a=rand()%(p-1)+1,temp=s;
        long long mod=modulo(a,temp,p);
        while(temp!=p-1 && mod!=1 && mod!=p-1){
            mod=mulmod(mod,mod,p);
            temp *= 2;
        }
        if(mod!=p-1 && temp%2==0){
            return false;
        }
    }
    return true;
}

int main(int argc, char* argv[])
{

    int input ;
    cin>>input;
    input++;
    int i = 0;

    if(input%2==0)
        i = input+1;
    else i = input;

    for(;i<2*input;i+=2) 
        if(Miller(i,20)) 
            break;
    cout<<i<<endl;

    return 0;
}