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: Accepted)

//#include <bits/stdc++.h>
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <cmath>
#include <stdio.h>
#include <complex>
#include <iomanip>
#include <queue>
#include <bitset>
#include <numeric>
//定数
#define MOD 1000000007
//マクロ
#define rep(i,n)    for(int (i)=0;(i)<(n);(i)++)
#define reprev(i,n) for(int (i)=(n-1);(i)>=0;(i)--)
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

#define vsort(v) sort((v).begin(), (v).end());

typedef unsigned long long int ull;
typedef long long lint;

using namespace std;



bool IsPrime(int num)
{
    if (num < 2)           return false;
    else if (num == 2)     return true;
    else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

    double sqrtNum = sqrt(num);
    for (int i = 3; i <= sqrtNum; i += 2)
    {
        if (num % i == 0)
        {
            // 素数ではない
            return false;
        }
    }

    // 素数である
    return true;
}

int main(){
    int X;

    //入力
    cin >> X;
    int i=X;
    while(!IsPrime(i)) i++;
    cout << i << endl;
    

    return 0;
}