Hull Sum

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Chef prepares the following problem for a programming contest:
You are given N points with integer coordinates (1000 ≤ x_{i}, y_{i} ≤ 1000). No two points collide and no three points are collinear. For each of the 2^{N}1 nonempty subsets of points, find the size (the number of points) of its convex hull. Print the sum of those sizes.
Chef has already prepared some tests for his problem, including a test with the maximum possible answer. He now needs a test with a quite small answer (but not necessarily the minimum possible one). For given N, your task is to find any valid set of N points for which the answer doesn't exceed 4 * 10^{15} (4,000,000,000,000,000).
Input
The only line of the input contains a single integer N, denoting the number of points.
Output
Print exactly N lines. The ith line should contain two integers x_{i} and y_{i}, denoting coordinates of the ith point.
The printed set of points must satisfy all the given conditions. At least one such set exists for every N allowed by the constraints given below.
Note: Remember that you don't print N in the first line.
Constraints
 2 ≤ N ≤ 50
Example
Input: 5 Output: 150 150 150 150 150 150 150 150 11 13
Explanation
Let's analyze sizes of convex hulls for the provided output. There are 2^{5}1 nonempty sets of points.
 There are 5 sets of points that consist of only 1 point each. The convex hull of each of those sets has size 1.
 Similarly, 10 sets of points consist of 2 points each, and the size of each of their convex hulls is 2.
 Similarly, there are 10 sets with 3 points each, and the size of each of their convex hulls is 3.
 The set with points (150, 150), (150, 150), (150, 150), (11, 13) has a convex hull of size 3. You can note that the point (11, 13) is inside the triangle formed by the other three points.
 The set with points (150, 150), (150, 150), (150, 150), (11, 13) has a convex hull of size 3 as well.
 There are 3 other sets with 4 points, and for each of them the convex hull has size 4.
 Finally, the set with all 5 points has a convex hull of size 4.
The sum of sizes of convex hulls is (5 * 1) + (10 * 2) + (10 * 3) + 3 + 3 + (3 * 4) + 4 = 5 + 20 + 30 + 6 + 12 + 4 = 77, which obviously doesn't exceed 4 * 10^{15}.
Author:  errichto 
Tags  errichto 
Date Added:  20062017 
Time Limit:  1 sec 
Source Limit:  50000 Bytes 
Languages:  C, CPP14, JAVA, PYTH, PYTH 3.5, PYPY, CS2, PAS fpc, PAS gpc, RUBY, PHP, GO, NODEJS, HASK, SCALA, D, PERL, FORT, WSPC, ADA, CAML, ICK, BF, ASM, CLPS, PRLG, ICON, SCM qobi, PIKE, ST, NICE, LUA, BASH, NEM, LISP sbcl, LISP clisp, SCM guile, JS, ERL, TCL, PERL6, TEXT, SCM chicken, CLOJ, FS 
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