#P1067C. Knights

    ID: 4987 Type: RemoteJudge 1000ms 256MiB Tried: 0 Accepted: 0 Difficulty: 10 Uploaded By: Tags>constructive algorithms*2600

Knights

Description

Ivan places knights on infinite chessboard. Initially there are nn knights. If there is free cell which is under attack of at least 44 knights then he places new knight in this cell. Ivan repeats this until there are no such free cells. One can prove that this process is finite. One can also prove that position in the end does not depend on the order in which new knights are placed.

Ivan asked you to find initial placement of exactly nn knights such that in the end there will be at least n210\lfloor \frac{n^{2}}{10} \rfloor knights.

The only line of input contains one integer nn (1n1031 \le n \le 10^{3}) — number of knights in the initial placement.

Print nn lines. Each line should contain 22 numbers xix_{i} and yiy_{i} (109xi,  yi109-10^{9} \le x_{i}, \,\, y_{i} \le 10^{9}) — coordinates of ii-th knight. For all iji \ne j, (xi,  yi)(xj,  yj)(x_{i}, \,\, y_{i}) \ne (x_{j}, \,\, y_{j}) should hold. In other words, all knights should be in different cells.

It is guaranteed that the solution exists.

Input

The only line of input contains one integer nn (1n1031 \le n \le 10^{3}) — number of knights in the initial placement.

Output

Print nn lines. Each line should contain 22 numbers xix_{i} and yiy_{i} (109xi,  yi109-10^{9} \le x_{i}, \,\, y_{i} \le 10^{9}) — coordinates of ii-th knight. For all iji \ne j, (xi,  yi)(xj,  yj)(x_{i}, \,\, y_{i}) \ne (x_{j}, \,\, y_{j}) should hold. In other words, all knights should be in different cells.

It is guaranteed that the solution exists.

Sample Input 1

4

Sample Output 1

1 1
3 1
1 5
4 4

Sample Input 2

7

Sample Output 2

2 1
1 2
4 1
5 2
2 6
5 7
6 6

Note

Let's look at second example:

Green zeroes are initial knights. Cell (3,  3)(3, \,\, 3) is under attack of 44 knights in cells (1,  2)(1, \,\, 2), (2,  1)(2, \,\, 1), (4,  1)(4, \,\, 1) and (5,  2)(5, \,\, 2), therefore Ivan will place a knight in this cell. Cell (4,  5)(4, \,\, 5) is initially attacked by only 33 knights in cells (2,  6)(2, \,\, 6), (5,  7)(5, \,\, 7) and (6,  6)(6, \,\, 6). But new knight in cell (3,  3)(3, \,\, 3) also attacks cell (4,  5)(4, \,\, 5), now it is attacked by 44 knights and Ivan will place another knight in this cell. There are no more free cells which are attacked by 44 or more knights, so the process stops. There are 99 knights in the end, which is not less than 7210=4\lfloor \frac{7^{2}}{10} \rfloor = 4.