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>

typedef long long LL;

#define FOR(i, a, b) for (int i = (a), i##_END_ = (b); i <= i##_END_; i++)
#define DNF(i, a, b) for (int i = (a), i##_END_ = (b); i >= i##_END_; i--)

template <typename Tp> void in(Tp &x) {
	char ch = getchar(), f = 1; x = 0;
	while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
	if (ch == '-') ch = getchar(), f = -1;
	while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
	x *= f;
}

template <typename Tp> void out(Tp x) {
	if (x > 9) out(x / 10);
	putchar(x % 10 + '0');
	return;
}

template <typename Tp> Tp Max(const Tp &x, const Tp &y) {return x > y ? x : y;}
template <typename Tp> Tp Min(const Tp &x, const Tp &y) {return x < y ? x : y;}
template <typename Tp> bool chkmax(Tp &x, Tp y) {return x >= y ? 0 : (x=y, 1);}
template <typename Tp> bool chkmin(Tp &x, Tp y) {return x <= y ? 0 : (x=y, 1);}

int n;
int ans[30010], sum;

void print() {
	FOR(i, 1, ans[0]) printf("%d ", ans[i]);
	putchar(10);
	exit(0);
}

int main()
{
	in(n);

	if (n == 3) {puts("2 5 63"); return 0;}

	while (true) {
		ans[0] = 0; sum = 0;
		FOR(i, 1, 30000) if (ans[0] < n && rand() % 5 < 5 && (i % 2 == 0 || i % 3 == 0)) {
			if (ans[0] == n - 1 && (sum + i) % 6 == 0) {
				ans[++ans[0]] = i; print();
			}
			else if (ans[0] != n - 1) {
				ans[++ans[0]] = i; sum += i;
			}
		}
	}

	return 0;
}