You are a paparazzi working in Manhattan.

Manhattan has $r$ south-to-north streets, denoted by numbers $1, 2,\ldots, r$ in order from west to east, and $r$ west-to-east streets, denoted by numbers $1,2,\ldots,r$ in order from south to north. Each of the $r$ south-to-north streets intersects each of the $r$ west-to-east streets; the intersection between the $x$-th south-to-north street and the $y$-th west-to-east street is denoted by $(x, y)$. In order to move from the intersection $(x,y)$ to the intersection $(x', y')$ you need $|x-x'|+|y-y'|$ minutes.

You know about the presence of $n$ celebrities in the city and you want to take photos of as many of them as possible. More precisely, for each $i=1,\dots, n$, you know that the $i$-th celebrity will be at the intersection $(x_i, y_i)$ in exactly $t_i$ minutes from now (and he will stay there for a very short time, so you may take a photo of him only if at the $t_i$-th minute from now you are at the intersection $(x_i, y_i)$). You are very good at your job, so you are able to take photos instantaneously. You know that $t_i \lt t_{i+1}$ for any $i=1,2,\ldots, n-1$.

Currently you are at your office, which is located at the intersection $(1, 1)$. If you plan your working day optimally, what is the maximum number of celebrities you can take a photo of?

Input

The first line of the input contains two positive integers $r, n$ ($1\le r\le 500$, $1\le n\le 100,000$) – the number of south-to-north/west-to-east streets and the number of celebrities.

Then $n$ lines follow, each describing the appearance of a celebrity. The $i$-th of these lines contains $3$ positive integers $t_i, x_i, y_i$ ($1\le t_i\le 1,000,000$, $1\le x_i, y_i\le r$)  — denoting that the $i$-th celebrity will appear at the intersection $(x_i, y_i)$ in $t_i$ minutes from now.

It is guaranteed that $t_i \lt t_{i+1}$ for any $i=1,2,\ldots, n-1$.

Output

Print a single integer, the maximum number of celebrities you can take a photo of.

Note

Explanation of the first testcase: There is only one celebrity in the city, and he will be at intersection $(6,8)$ exactly $11$ minutes after the beginning of the working day. Since you are initially at $(1,1)$ and you need $|1-6|+|1-8|=5+7=12$ minutes to reach $(6,8)$ you cannot take a photo of the celebrity. Thus you cannot get any photo and the answer is $0$.

Explanation of the second testcase: One way to take $4$ photos (which is the maximum possible) is to take photos of celebrities with indexes $3, 5, 7, 9$ (see the image for a visualization of the strategy):

• To move from the office at $(1,1)$ to the intersection $(5,5)$ you need $|1-5|+|1-5|=4+4=8$ minutes, so you arrive at minute $8$ and you are just in time to take a photo of celebrity $3$.
• Then, just after you have taken a photo of celebrity $3$, you move toward the intersection $(4,4)$. You need $|5-4|+|5-4|=1+1=2$ minutes to go there, so you arrive at minute $8+2=10$ and you wait until minute $12$, when celebrity $5$ appears.
• Then, just after you have taken a photo of celebrity $5$, you go to the intersection $(6,6)$. You need $|4-6|+|4-6|=2+2=4$ minutes to go there, so you arrive at minute $12+4=16$ and you wait until minute $17$, when celebrity $7$ appears.
• Then, just after you have taken a photo of celebrity $7$, you go to the intersection $(5,4)$. You need $|6-5|+|6-4|=1+2=3$ minutes to go there, so you arrive at minute $17+3=20$ and you wait until minute $21$ to take a photo of celebrity $9$.

Explanation of the third testcase: The only way to take $1$ photo (which is the maximum possible) is to take a photo of the celebrity with index $1$ (since $|2-1|+|1-1|=1$, you can be at intersection $(2,1)$ after exactly one minute, hence you are just in time to take a photo of celebrity $1$).

Explanation of the fourth testcase: One way to take $3$ photos (which is the maximum possible) is to take photos of celebrities with indexes $3, 8, 10$:

• To move from the office at $(1,1)$ to the intersection $(101,145)$ you need $|1-101|+|1-145|=100+144=244$ minutes, so you can manage to be there when the celebrity $3$ appears (at minute $341$).
• Then, just after you have taken a photo of celebrity $3$, you move toward the intersection $(149,379)$. You need $|101-149|+|145-379|=282$ minutes to go there, so you arrive at minute $341+282=623$ and you wait until minute $682$, when celebrity $8$ appears.
• Then, just after you have taken a photo of celebrity $8$, you go to the intersection $(157,386)$. You need $|149-157|+|379-386|=8+7=15$ minutes to go there, so you arrive at minute $682+15=697$ and you wait until minute $855$ to take a photo of celebrity $10$.