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D. Petya and His Friends

time limit per test

2 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputLittle Petya has a birthday soon. Due this wonderful event, Petya's friends decided to give him sweets. The total number of Petya's friends equals to *n*.

Let us remind you the definition of the greatest common divisor: *GCD*(*a* _{1}, ..., *a* _{ k}) = *d*, where *d* represents such a maximal positive number that each *a* _{ i} (1 ≤ *i* ≤ *k*) is evenly divisible by *d*. At that, we assume that all *a* _{ i}'s are greater than zero.

Knowing that Petya is keen on programming, his friends has agreed beforehand that the 1-st friend gives *a* _{1} sweets, the 2-nd one gives *a* _{2} sweets, ..., the *n*-th one gives *a* _{ n} sweets. At the same time, for any *i* and *j* (1 ≤ *i*, *j* ≤ *n*) they want the *GCD*(*a* _{ i}, *a* _{ j}) not to be equal to 1. However, they also want the following condition to be satisfied: *GCD*(*a* _{1}, *a* _{2}, ..., *a* _{ n}) = 1. One more: all the *a* _{ i} should be distinct.

Help the friends to choose the suitable numbers *a* _{1}, ..., *a* _{ n}.

Input

The first line contains an integer *n* (2 ≤ *n* ≤ 50).

Output

If there is no answer, print "-1" without quotes. Otherwise print a set of *n* distinct positive numbers *a* _{1}, *a* _{2}, ..., *a* _{ n}. Each line must contain one number. Each number must consist of not more than 100 digits, and must not contain any leading zeros. If there are several solutions to that problem, print any of them.

Do not forget, please, that all of the following conditions must be true:

- For every
*i*and*j*(1 ≤*i*,*j*≤*n*):*GCD*(*a*_{ i},*a*_{ j}) ≠ 1 -
*GCD*(*a*_{1},*a*_{2}, ...,*a*_{ n}) = 1 - For every
*i*and*j*(1 ≤*i*,*j*≤*n*,*i*≠*j*):*a*_{ i}≠*a*_{ j}

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Examples

Input

3

Output

99

55

11115

Input

4

Output

385

360

792

8360

Codeforces (c) Copyright 2010-2020 Mike Mirzayanov

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