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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <cassert>
#include <algorithm>
#include <sstream>
#include <set>
using namespace std;
typedef long long ll;


long long answer[10] = {};

set< pair<int,int>> memo;

int main(){
	int H, W, N;
	cin >> H >> W >> N;
	
	answer[0] = (H - 2) * (W - 2);
	
	set< pair<int,int> > S;
	for(int i = 0 ; i < N ; i++){
		int a, b;
		cin >> a >> b;
		--a, --b;
		memo.insert({a, b});
		for(int j = 0 ; j < 3 ; j++){
			for(int k = 0 ; k < 3 ; k++){
				int y = a - j;
				int x = b - k;
				if( x < 0 || y < 0 ) continue;
				if( x + 3 > W || y + 3 > H ) continue;
				S.insert({y, x});
			}
		}
	}
	
	for( auto p : S ) {
		int y = p.first;
		int x = p.second;
		int cnt = 0;
		for(int j = 0 ; j < 3 ; j++){
			for(int k = 0 ; k < 3 ; k++){
				int ty = p.first + j;
				int tx = p.second + k;
				if( memo.count({ty, tx}) ) cnt++; 
			}
		}
		answer[0]--;
		answer[cnt]++;
	}
	
	for(int i = 0 ; i < 10 ; i++){
		cout << answer[i] << endl;
	}
}