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 Python (3.4.3) (judge verdict: Accepted)

# https://atcoder.jp/contests/abc096/tasks/abc096_d
# 制約は小さい
# まず 55555以下の素数をエラトステネスの篩であぶり出す

# 55555以下の素数からN個選ぶ。N個の中からどの5つの素数を選んでも合成数となるのが良い

# 性質を列挙
# 5の倍数+1の素数を5a+1とする #aはなんでも
# このとき、5つの和は 5a+1 + 5b+1 + ... + 5d+1 で5で割り切れる合成数となる
# よって5の倍数+1となるような素数を探して列挙すれば良い


def ret_erators(N: int):
    # エラトステネスの篩
    is_prime = [True] * (N + 1)

    # 0と1は素数ではない
    is_prime[0] = False
    is_prime[1] = False

    for i in range(2, int(N**0.5) + 1):
        if is_prime[i]:
            j = i * 2  # iの倍数は素数ではない
            while j < N + 1:
                is_prime[j] = False
                j += i
    return is_prime


is_prime = ret_erators(55555)
# print(len(is_prime))
# print(sum(is_prime))
candi = []
for i in range(55555):
    n = i * 5 + 1
    if n > 55555:
        break
    if is_prime[n]:
        candi.append(n)


print(*candi[:int(input())])