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 9.2.1) (judge verdict: Accepted)

//g++ -std=c++17 -Wshadow -Wall -fsanitize=address -fsanitize=undefined -D_GLIBCXX_DEBUG -g PROGRAM -o OUTPUT -DLOCAL -lasan -lubsan
#include <bits/stdc++.h>

#define ll long long
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define eps 1e-6
#define pi 3.141592653589793
using namespace std;

typedef vector<int> VII;
typedef vector<ll> VLL;
typedef pair<int, int> PII;
typedef vector<pair<int, int>> VPII;
typedef vector<pair<int, PII>> VPPI;
const int MOD = 1e9 + 7;
const int INF = 1e9;


inline ll modulo(ll a, ll m)
{
    return (a % m + m) % m;
}

inline ll modInverse(ll a, ll m)
{
    assert(__gcd(a, m) == 1);
    ll m0 = m;
    ll y = 0, x = 1;

    if (m == 1)
        return 0;

    while (a > 1)
    {
        ll q = a / m;
        ll t = m;

        m = a % m, a = t;
        t = y;

        y = x - q * y;
        x = t;
    }

    if (x < 0)
        x += m0;

    return x;
}

inline ll modPow(ll x, ll y, ll m)
{ //x^y % m
    if (y == 0)
        return 1LL;
    else if (y == 1)
        return x;
    else
    {
        ll ans = modPow(x, y / 2, m) % m;
        if (y & 1)
        {
            return (((ans * ans) % m) * x) % m;
        }
        else
        {
            return (ans * ans) % m;
        }
    }
}

vector<string> split(char d,string s)
{
    stringstream ss(s);
    string t;
    vector<string> words;
    while(getline(ss,t,d))
    {
        words.pb(t);
    }

    return words;
}

int main(int argc, char *argv[])
{
    
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    VII primes;
    vector<bool> vis(1e6, false);
    int l = 1e6;
    for (int i = 2; i <= l;++i)
    {
        if(vis[i])
            continue;

        primes.pb(i);
        for (int j = i + i; j <= l;j+=i)
            vis[j] = true;
    }

    int n;
    cin >> n;
    cout<<*lower_bound(all(primes), n);
    return 0;
}