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>
using namespace std;

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)

using ll = long long;
using ld = long double;
ll INF = LLONG_MAX;

using vi = vector<int>;
using vll = vector<ll>;
using pii = pair<int, int>;

namespace output {
	void pr(int x) { cout << x; }
	void pr(long x) { cout << x; }
	void pr(ll x) { cout << x; }
	void pr(unsigned x) { cout << x; }
	void pr(unsigned long x) { cout << x; }
	void pr(unsigned long long x) { cout << x; }
	void pr(float x) { cout << x; }
	void pr(double x) { cout << x; }
	void pr(ld x) { cout << x; }
	void pr(char x) { cout << x; }
	void pr(const char* x) { cout << x; }
	void pr(const string& x) { cout << x; }
	void pr(bool x) { pr(x ? "true" : "false"); }
	template<class T> void pr(const complex<T>& x) { cout << x; }
	
	template<class T1, class T2> void pr(const pair<T1,T2>& x);
	template<class T> void pr(const T& x);
	
	template<class T, class... Ts> void pr(const T& t, const Ts&... ts) { 
		pr(t); pr(ts...); 
	}
	template<class T1, class T2> void pr(const pair<T1,T2>& x) { 
		pr("{",x.f,", ",x.s,"}"); 
	}
	template<class T> void pr(const T& x) { 
		pr("{"); // const iterator needed for vector<bool>
		bool fst = 1; for (const auto& a: x) pr(!fst?", ":"",a), fst = 0; 
		pr("}");
	}
	
	void print() { pr("\n"); } // print w/ spaces
	template<class T, class... Ts> void print(const T& t, const Ts&... ts) { 
		pr(t); if (sizeof...(ts)) pr(" "); print(ts...); 
	}
}

using namespace output;

int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int N; cin >> N;
	
	vi ans;

	int pos = 0;
	while (ans.size() < N) {
		ans.push_back(pos + 2);
		ans.push_back(pos + 3);
		ans.push_back(pos + 4);
		ans.push_back(pos + 6);
		pos += 6;
	}
	
	int addTo = 3;
	while (ans.size() > N) {
		ans[addTo--] += ans[ans.size() - 1];
		ans.pop_back();
	}

	if (N <= 4) {
		F0R(i, N) ans[i] = 2 * (i + 1);
	}

	for (int x : ans) {
		cout << x << " ";
	}
	cout << endl;
}