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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.Scanner;
import java.util.Map; 
import java.util.HashMap;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int h = sc.nextInt();
        int w = sc.nextInt();
        int n = sc.nextInt();
        
        Map<Integer, Map<Integer, Integer>> row = new HashMap<Integer, Map<Integer, Integer>>();
        for (int i = 0; i < n; i++) {
            int a = sc.nextInt() - 1;
            int b = sc.nextInt() - 1;
            for (int j = 0; j < 3; j++) {
                for (int k = 0; k < 3; k++) {
                    if (a - j < 0 || b - k < 0) continue;
                    else {
                        if (row.containsKey(a - j)) {
                            if (row.get(a - j).containsKey(b - k)) {
                                row.get(a - j).put(b - k, row.get(a - j).get(b - k) + 1);
                            } else {
                                row.get(a - j).put(b - k, 1);
                            }
                        } else {
                            Map<Integer, Integer> column = new HashMap<Integer, Integer>();
                            column.put(b - k, 1);
                            row.put(a - j, column);
                        }
                    }
                }
            }
        }
        sc.close();

        long count[] = new long[10];
        for (int i = 0; i < 10; i++) {
            count[i] = 0;
        }

        for (int i = 0; i < h - 2; i++) {
            if (row.containsKey(i)) {
                for (int j = 0; j < w - 2; j++) {
                    if (row.get(i).containsKey(j)) {
                        count[row.get(i).get(j)]++;
                    } else count[0]++;
                }
            } else {
                count[0] += (w - 2);
                continue;
            }
        }

        for (int i = 0; i < 10; i++) {
            System.out.println(count[i]);
        }
    }
}