Problem:
Score : 400 points

Problem Statement

We have a grid with H rows and W columns. At first, all cells were painted white.

Snuke painted N of these cells. The i-th ( 1 \leq i \leq N ) cell he painted is the cell at the a_i-th row and b_i-th column.

Compute the following:

For each integer j ( 0 \leq j \leq 9 ), how many subrectangles of size 3×3 of the grid contains exactly j black cells, after Snuke painted N cells?

Constraints

3 \leq H \leq 10^9

3 \leq W \leq 10^9

0 \leq N \leq min(10^5,H×W)

1 \leq a_i \leq H (1 \leq i \leq N)

1 \leq b_i \leq W (1 \leq i \leq N)

(a_i, b_i) \neq (a_j, b_j) (i \neq j)

Input

The input is given from Standard Input in the following format:

H W N
a_1 b_1
:
a_N b_N

Output

Print 10 lines.
The (j+1)-th ( 0 \leq j \leq 9 ) line should contain the number of the subrectangles of size 3×3 of the grid that contains exactly j black cells.

Sample Input 1

4 5 8
1 1
1 4
1 5
2 3
3 1
3 2
3 4
4 4

Sample Output 1

0
0
0
2
4
0
0
0
0
0

There are six subrectangles of size 3×3. Two of them contain three black cells each, and the remaining four contain four black cells each.

Sample Input 2

10 10 20
1 1
1 4
1 9
2 5
3 10
4 2
4 7
5 9
6 4
6 6
6 7
7 1
7 3
7 7
8 1
8 5
8 10
9 2
10 4
10 9

Sample Output 2

4
26
22
10
2
0
0
0
0
0

Sample Input 3

1000000000 1000000000 0

Sample Output 3

999999996000000004
0
0
0
0
0
0
0
0
0

Solution in Rust (1.15.1) (judge verdict: Accepted)

use std::io::{ self, Write };
use std::str::FromStr;
use std::cmp::{ min, max };
use std::collections::{ BinaryHeap, VecDeque, HashSet };

macro_rules! trace {
    ($var:expr) => ({
        let _ = writeln!(&mut std::io::stderr(), ">>> {} = {:?}", stringify!($var), $var);
    })
}
macro_rules! swap { ($a:expr, $b:expr) => ({ let t = $b; $b = $a; $a = t; }) }

fn main() {
    let mut sc = Scanner::new();

    let h: i32 = sc.cin();
    let w: i32 = sc.cin();

    let mut ps = vec![];

    let n: usize = sc.cin();
    for i in 0..n {
        let a: i32 = sc.cin();
        let b: i32 = sc.cin();
        ps.push((a, b));
    }

    let mut qs = HashSet::new();
    let mut bs = HashSet::new();

    for &p in ps.iter() {
        let (a, b) = p;
        bs.insert((a, b));
        for i in (a-1)..(a+2) {
            for j in (b-1)..(b+2) {
                qs.insert((i, j));
            }
        }
    }

    let mut counts = [0 as u64; 10];

    for &p in &qs {
        let (a, b) = p;
        if a <= 1 || a >= h || b <= 1 || b >= w { continue }
        let mut count = 0;
        for i in (a-1)..(a+2) {
            for j in (b-1)..(b+2) {
                if bs.contains(&(i, j)) {
                    count += 1
                }
            }
        }
        // trace!((p, count));
        counts[count] += 1;
    }

    counts[0] = (h - 2) as u64 * (w - 2) as u64;
    for i in 1..10 { counts[0] -= counts[i]; }

    for i in 0..10 {
        println!("{}", counts[i]);
    }
}

#[allow(dead_code)]
struct Scanner { stdin: io::Stdin, buffer: VecDeque<String>, }
#[allow(dead_code)]
impl Scanner {
    fn new() -> Scanner { Scanner { stdin: io::stdin(), buffer: VecDeque::new() } }
    fn reserve(&mut self) {
        while self.buffer.len() == 0 {
            let mut line = String::new();
            let _ = self.stdin.read_line(&mut line);
            for w in line.split_whitespace() {
                self.buffer.push_back(String::from(w));
            }
        }
    }
    fn cin<T: FromStr>(&mut self) -> T {
        self.reserve();
        match self.buffer.pop_front().unwrap().parse::<T>() {
            Ok(a) => a,
            Err(_) => panic!("parse err")
        }
    }
    fn get_char(&mut self) -> char {
        self.reserve();
        let head = self.buffer[0].chars().nth(0).unwrap();
        let tail = String::from( &self.buffer[0][1..] );
        if tail.len()>0 { self.buffer[0]=tail } else { self.buffer.pop_front(); }
        head
    }
}
struct PowerPermutation { n: usize, m: usize, ar: Vec<usize> }
impl PowerPermutation {
    fn new(n: usize, m: usize) -> PowerPermutation {
        let ar = vec![0; m];
        PowerPermutation { n: n, m: m, ar: ar }
    }
}
impl Iterator for PowerPermutation {
    type Item = Vec<usize>;
    fn next(&mut self) -> Option<Vec<usize>> {
        if self.ar[self.m-1] >= self.n { return None }
        let r = self.ar.clone();
        self.ar[0] += 1;
        for i in 0..self.m-1 {
            if self.ar[i] == self.n {
                self.ar[i] = 0;
                self.ar[i + 1] += 1;
            } else {
                break;
            }
        }
        return Some(r);
    }
}