## 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 non-empty 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 this 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). Forgiven **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 i-th line should contain two integers x_{i} and y_{i}, denoting coordinates of the i-th 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 non-empty 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}.