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>
#define rep(i,n) for (int i = 0; i < (n); i++)
#define repe(i,n) for (int i = 0; i <= (n); ++i)
#define repe1(i,n) for (int i = 0; i <= (n); ++i)
#define all(x) (x).begin(),(x).end()
#define pb(x) push_back(x)
using namespace std;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
typedef long long ll;
const int INF = 1e9;
const ll mod = 1000000007;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a * b / gcd(a, b);}

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;
    cin >> x;
    for(int i=x; x<1000000; i++) {
        if (IsPrime(i)) {
            cout << i << "\n";
            break;
        }
    }
}