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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

#define DUMP(x) cerr << #x << "=" << x << endl
#define DUMP2(x, y) cerr<<"("<<#x<<", "<<#y<<") = ("<<x<<", "<<y<<")"<< endl
#define BINARY(x) static_cast<bitset<16> >(x)

#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define REP(i,m,n) for (int i=m;i<(int)(n);i++)

#define in_range(x, y, w, h) (0<=(int)(x) && (int)(x)<(int)(w) && 0<=(int)(y) && (int)(y)<(int)(h))
#define ALL(a) (a).begin(),(a).end()

typedef long long ll;
const int INF   = 1e9;
const ll  INFLL = 1e18;
typedef pair<int, int> PII;
typedef vector<vector<int>> Graph;
int dx[4]={0, -1, 1, 0}, dy[4]={-1, 0, 0, 1};

void solve(int N)
{
    if (N == 3) {
        cout << "2 5 63" << endl;
        return;
    }
    set<ll> S;
    
    ll sum = 0, num = -1;
    for (int i=2; S.size() < N-1; i++) {
        if (i%2 == 0 || i%3 == 0) {
            cout << i << " ";
            num = i;
            sum += i;
            S.insert(num);
        }
    }

    DUMP(sum);
    ll last_sum = (sum / 6 + 1) * 6;

    while (true) {
        ll tmp = last_sum - sum;
        if (!S.count(tmp) && (tmp%2 == 0 || tmp%3 == 0)) break;
        last_sum += 6;
    }
    cout << last_sum - sum << endl;
    DUMP(last_sum);

    assert(last_sum - sum <= 30000);
}

int main()
{
    ios::sync_with_stdio(false);
    int N; cin >> N;
    solve(N);
}