Problem:
Score : 600 points

Problem Statement

Nagase is a top student in high school. One day, she's analyzing some properties of special sets of positive integers.

She thinks that a set S = \{a_{1}, a_{2}, ..., a_{N}\} of distinct positive integers is called special if for all 1 \leq i \leq N, the gcd (greatest common divisor) of a_{i} and the sum of the remaining elements of S is not 1.

Nagase wants to find a special set of size N. However, this task is too easy, so she decided to ramp up the difficulty. Nagase challenges you to find a special set of size N such that the gcd of all elements are 1 and the elements of the set does not exceed 30000.

Constraints

3 \leq N \leq 20000

Input

Input is given from Standard Input in the following format:

N

Output

Output N space-separated integers, denoting the elements of the set S. S must satisfy the following conditions :

The elements must be distinct positive integers not exceeding 30000.

The gcd of all elements of S is 1, i.e. there does not exist an integer d > 1 that divides all elements of S.

S is a special set.

If there are multiple solutions, you may output any of them. The elements of S may be printed in any order. It is guaranteed that at least one solution exist under the given contraints.

Sample Input 1

3

Sample Output 1

2 5 63

\{2, 5, 63\} is special because gcd(2, 5 + 63) = 2, gcd(5, 2 + 63) = 5, gcd(63, 2 + 5) = 7. Also, gcd(2, 5, 63) = 1. Thus, this set satisfies all the criteria.

Note that \{2, 4, 6\} is not a valid solution because gcd(2, 4, 6) = 2 > 1.

Sample Input 2

4

Sample Output 2

2 5 20 63

\{2, 5, 20, 63\} is special because gcd(2, 5 + 20 + 63) = 2, gcd(5, 2 + 20 + 63) = 5, gcd(20, 2 + 5 + 63) = 10, gcd(63, 2 + 5 + 20) = 9. Also, gcd(2, 5, 20, 63) = 1. Thus, this set satisfies all the criteria.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define SORT(x) sort((x).begin(), (x).end())
#define ALL(x) (x).begin(), (x).end()
#define rep(i, n) for (ll i = 0; i < n; i++)
#define reps(i, m, n) for (ll i = m; i < n; i++)
#define repr(i, m, n) for (ll i = m; i >= n; i--)
#define de(x) cout << #x << "=" << x << endl;
template <class T>
bool maxi(T &a, const T &b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
bool mini(T &a, const T &b) {
    if (b < a) {
        a = b;
        return 1;
    }
    return 0;
}
#define INF2 1000000000000000037
#define INF  1000000007
#define MOD  1000000007
#define PI (acos(-1))
using namespace std;
using ll = long long;
using ld = long double;
using P  = pair<ll,ll>;
//--GLOBAL---------------------------------

//--MAIN-----------------------------------
void Main() {
    ll N; cin>>N;
    vector<ll> ans(N);
    ll ct=1;
    ll sum=2;
    ans[0]=2;
    ll buf=-1;
    reps(i,3,27500){
        if(i%2==0){
            if(ct==N-2) continue;
            if(buf==-1){
                buf=i;
            }
            else{
                ans[ct++]=buf;
                ans[ct++]=i;
                buf=-1;
            }
        }
        else if(i%3==0 or i%5==0 or i%7==0 or i%11==0){
            ans[ct++]=i;
            sum+=i;
        }
        if(ct==N-1) break;
        if(i==27499) assert(0);
    }
    repr(i,30000,1){
        if(i%2==0 or i%3==0 or i%5==0 or i%7==0 or i%11==0){
            if((sum+i)%2310==0){
                ans[N-1]=i; break;
            }
        }
    }
    rep(i,N){
        cout<< ans[i] <<" ";
    }
    cout<<"\n";
}
//--START----------------------------------
int main() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    Main();
}
//-----------------------------------------