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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <sstream>
#include <queue>
#include <deque>
#include <bitset>
#include <iterator>
#include <list>
#include <stack>
#include <map>
#include <set>
#include <functional>
#include <numeric>
#include <utility>
#include <limits>
#include <ctime>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cassert>
#include <unordered_map>
#include <unordered_set>

using namespace std;

typedef long long lli;
typedef long double ld;
typedef pair<lli, lli > plli;
typedef vector<lli> vlli;
typedef vector<plli > vplli;

#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);
#define lcm(a, b) ((a * b) / gcd(a, b))
#define sqr(x) (x)*(x)
#define all(a) a.begin(),a.end()
#define UN(v) sort(all(v)), v.resize(unique(all(v))-v.begin())

#define endl '\n'

const long long INF = 100000000000000000;
const long long MOD = 1000000007;
const long long MAXN = 100005;
const double PI = acos(-1);

lli dx[]={0,0,-1,1,-1,-1,1,1};
lli dy[]={1,-1,0,0,1,-1,-1,1};

lli gcd(lli a , lli b) {
    if(b == 0) return a;
    a %= b;
    return gcd(b,a);
}

lli nums[MAXN + 5];
vlli primes;

void sieve() {
    memset(nums, 1, sizeof(nums));

    for (int i = 2; i * i <= MAXN; i++) {
        if (nums[i]) {
            for (int j = 2 * i; j <= MAXN; j += i) {
                nums[j] = 0;
            }
        }
    }

    for (int i = 2; i <= MAXN; i++) {
        if (nums[i]) {
//            cout << i << endl;
            primes.push_back(i);
        }
    }
}
void solve() {
    sieve();
    lli x;
    cin >> x;
    cout << *(lower_bound(all(primes), x)) << endl;
}

int main()
{
    IOS;
    int t = 1;
    //cin >> t;
    for (int i = 1; i <= t; i++) {
        //cout << "Case " << i << ": ";
        solve();
    }
    return 0;
}