Problem:
Score: 400 points

Problem Statement

Print a sequence a_1, a_2, ..., a_N whose length is N that satisfies the following conditions:

a_i (1 \leq i \leq N) is a prime number at most 55 555.

The values of a_1, a_2, ..., a_N are all different.

In every choice of five different integers from a_1, a_2, ..., a_N, the sum of those integers is a composite number.

If there are multiple such sequences, printing any of them is accepted.

Notes

An integer N not less than 2 is called a prime number if it cannot be divided evenly by any integers except 1 and N, and called a composite number otherwise.

Constraints

N is an integer between 5 and 55 (inclusive).

Input

Input is given from Standard Input in the following format:

N

Output

Print N numbers a_1, a_2, a_3, ..., a_N in a line, with spaces in between.

Sample Input 1

5

Sample Output 1

3 5 7 11 31

Let us see if this output actually satisfies the conditions.

First, 3, 5, 7, 11 and 31 are all different, and all of them are prime numbers.

The only way to choose five among them is to choose all of them, whose sum is a_1+a_2+a_3+a_4+a_5=57, which is a composite number.

There are also other possible outputs, such as 2 3 5 7 13, 11 13 17 19 31 and 7 11 5 31 3.

Sample Input 2

6

Sample Output 2

2 3 5 7 11 13

2, 3, 5, 7, 11, 13 are all different prime numbers.

2+3+5+7+11=28 is a composite number.

2+3+5+7+13=30 is a composite number.

2+3+5+11+13=34 is a composite number.

2+3+7+11+13=36 is a composite number.

2+5+7+11+13=38 is a composite number.

3+5+7+11+13=39 is a composite number.

Thus, the sequence 2 3 5 7 11 13 satisfies the conditions.

Sample Input 3

8

Sample Output 3

2 5 7 13 19 37 67 79

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

#include <stdio.h>
#include <iostream>
#include <cstdio>
#include <algorithm> 
#include <stdio.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cfloat>
#include <map>
#include <list>
#include <utility>
#include <set>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include <numeric>
#include <bitset>
#include <stdio.h>
#include <string.h>
#include <assert.h>
#include <random>

using namespace std;

static const double EPS = 1e-9;
template<class T> bool INRANGE(T x, T a, T b) { return a <= x && x <= b; }
template<class T> void amin(T &a, T v) { if (a > v) a = v; }
template<class T> void amax(T &a, T v) { if (a < v) a = v; }
int ROUND(double x) { return (int)(x + 0.5); }
bool ISINT(double x) { return fabs(ROUND(x) - x) <= EPS; }
bool ISEQUAL(double x, double y) { return fabs(x - y) <= EPS * max(1.0, max(fabs(x), fabs(y))); }
double SQSUM(double x, double y) { return x * x + y * y; }
template<class T>
vector<T> make_vector(int n, T t) {
	return vector<T>(n, t);
}
 
template<class ...Ts>
auto make_vector(int n, Ts ... ts) {
	return vector<decltype(make_vector(ts...))>(n, make_vector(ts...));
}

#define PI  (acos(-1))
#define ARRAY_NUM(a) (sizeof(a)/sizeof(a[0])) 
#define NG (-1)
#define BIG ((int)1e9+10)
#define BIGLL ((ll)4e18)
#define SZ(a) ((int)(a).size())
#define SQ(a) ((a)*(a))
#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define ALL(x) (x).begin(),(x).end()
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define POSSIBLE(n) cout << ((n) ? "POSSIBLE" : "IMPOSSIBLE"  ) << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible"  ) << endl

#define BitSet(arg,posn) ((arg) |= (1LL << (posn)))
#define BitClr(arg,posn) ((arg) &= ~(1LL << (posn)))
#define BitFlp(arg,posn) ((arg) ^= (1LL << (posn)))
#define IsBit(arg,posn) static_cast<bool>((arg) & (1LL << (posn)))

// #define DEBUG(x) cerr<<#x<<": "<<(x)<<endl
// #define DEBUG_VEC(v) cerr<<#v<<": ";REP(__i,(v).size())cerr<<((v)[__i])<<", ";cerr<<endl

typedef unsigned long long ull;
typedef long long ll;

struct in {
	int n;
	in() {}
	in(int n_) : n(n_) {};
	template <class T> operator T() {
		T ret;
		cin >> ret;
		return ret;
	}
	template <class T> operator vector<T>() {
		vector<T> ret(n);
		for (int i = 0; i < n; i++) cin >> ret[i];
		return ret;
	}
};

// modint: mod 計算を int を扱うように扱える構造体
template<int MOD>
struct ModInt {
	long long val;
	constexpr ModInt(long long v = 0) noexcept : val(v % MOD) {
		if (val < 0) v += MOD;
	}
	constexpr int getmod() { return MOD; }
	constexpr ModInt operator - () const noexcept {
		return val ? MOD - val : 0;
	}
	constexpr ModInt operator + (const ModInt& r) const noexcept { return ModInt(*this) += r; }
	constexpr ModInt operator - (const ModInt& r) const noexcept { return ModInt(*this) -= r; }
	constexpr ModInt operator * (const ModInt& r) const noexcept { return ModInt(*this) *= r; }
	constexpr ModInt operator / (const ModInt& r) const noexcept { return ModInt(*this) /= r; }
	constexpr ModInt& operator += (const ModInt& r) noexcept {
		val += r.val;
		if (val >= MOD) val -= MOD;
		return *this;
	}
	constexpr ModInt& operator -= (const ModInt& r) noexcept {
		val -= r.val;
		if (val < 0) val += MOD;
		return *this;
	}
	constexpr ModInt& operator *= (const ModInt& r) noexcept {
		val = val * r.val % MOD;
		return *this;
	}
	constexpr ModInt& operator /= (const ModInt& r) noexcept {
		long long a = r.val, b = MOD, u = 1, v = 0;
		while (b) {
			long long t = a / b;
			a -= t * b; swap(a, b);
			u -= t * v; swap(u, v);
		}
		val = val * u % MOD;
		if (val < 0) val += MOD;
		return *this;
	}
	constexpr bool operator == (const ModInt& r) const noexcept {
		return this->val == r.val;
	}
	constexpr bool operator != (const ModInt& r) const noexcept {
		return this->val != r.val;
	}

	friend ostream& operator << (ostream &os, const ModInt<MOD>& x) noexcept {
		return os << x.val;
	}
	friend istream& operator >> (istream &is, ModInt<MOD>& x) noexcept {
		return is >> x.val;
	}
	friend constexpr ModInt<MOD> modpow(const ModInt<MOD> &a, long long n) noexcept {
		if (n == 0) return 1;
		auto t = modpow(a, n / 2);
		t = t * t;
		if (n & 1) t = t * a;
		return t;
	}
};

// 二項係数ライブラリ
template<class T> struct BiCoef
{
	constexpr BiCoef() {}
	constexpr BiCoef(int n) noexcept : mFact(n, 1), mInv(n, 1), mFInv(n, 1)
	{
		Init(n);
	}

	constexpr void Init(int n) noexcept
	{
		mFact.assign(n, 1), mInv.assign(n, 1), mFInv.assign(n, 1);
		int factMod = mFact[0].getmod();
		for (int i = 2; i < n; i++) {
			mFact[i] = mFact[i - 1] * i;
			mInv[i] = -mInv[factMod%i] * (factMod / i);
			mFInv[i] = mFInv[i - 1] * mInv[i];
		}
	}

	constexpr T Fact(int n) const noexcept {
		if (n < 0) return 0;
		return mFact[n];
	}

	constexpr T Comb(int n, int k) const noexcept {
		if (n < k || n < 0 || k < 0) return 0;
		return mFact[n] * mFInv[k] * mFInv[n - k];
	}

	constexpr T Perm(int n, int k) const noexcept {
		if (n < k || n < 0 || k < 0) return 0;
		return mFact[n] * mFInv[k] * mFInv[n - k];
	}

	constexpr T Homo(int n, int r) const {
		if (n < 0 || r < 0) return 0;
		return r == 0 ? 1 : Comb(n + r - 1, r);
	}

	constexpr T inv(int n) const noexcept {
		if (n < 0) return 0;
		return mInv[n];
	}

	constexpr T finv(int n) const noexcept {
		if (n < 0) return 0;
		return mFInv[n];
	}

	vector<T> mFact;
	vector<T> mInv;
	vector<T> mFInv;
};

const int MOD = 1000000007;
using mint = ModInt<MOD>;
// BiCoef<mint> bc(1234567); // 1M * 3 * 8B(ll) = 24MB 

void make_primes(int num, vector<int>& primes)
{
	primes.clear();
	primes.reserve(num);
	vector<int> no_prime;
	no_prime.resize(num + 1);

	for (int i = 2; i <= num; i++)
	{
		if (!no_prime[i])
		{
			primes.push_back(i);
			for (int j = i + i; j <= num; j += i)
			{
				no_prime[j] = 1;
			}
		}
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	vector<int> primes;
	make_primes(55556, primes);

	int N = in();

	vector <int> ans;
	for (int i = 0; i < SZ(primes); ++i)
	{
		if (primes[i] % 5 == 1)
		{
			ans.push_back(primes[i]);
			if (SZ(ans) == N)
			{
				break;
			}
		}
	}

	for (int i = 0; i < SZ(ans); ++i)
	{
		if(i>0) 		printf(" ");
		printf("%d", ans[i]);


	}
	printf("\n");




	return 0;
}