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: Accepted)

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <math.h>
#include <string.h>
#include <string>
#include <queue>
#include <stack>
using namespace std;
typedef long long int ll;
map<pair<ll, ll> , int> visited;
pair<ll, ll> point[111111];
ll h, w, n;
ll ans[10];
int dy[9]={0,1,1,1,0,-1,-1,-1,0};
int dx[9]={1,1,0,-1,-1,-1,0,1,0};

int exist(ll u, ll v)
{
	if(lower_bound(point+1,point+n+1,make_pair(u,v))!=upper_bound(point+1,point+n+1,make_pair(u,v)))
	{
		return 1;
	}
	return 0;
}

int main(void)
{
	cin>>h>>w>>n;
	int i,j, k;
	ll u, v;
	ll tot=(h-2)*(w-2);
	for(i=1 ; i<=n ; i++)
	{
		scanf("%lld %lld",&u,&v);
		point[i]=make_pair(u,v);
	}
	sort(point+1,point+n+1);
	for(i=1 ; i<=n ; i++)
	{
		for(j=0 ; j<9 ; j++)
		{
			int cnt=0;
			ll y, x;
			y=point[i].first+dy[j];
			x=point[i].second+dx[j];
			if(visited.find(make_pair(y,x))==visited.end() && y>=2 && y<=h-1 && x>=2 && x<=w-1)
			{
				for(k=0 ; k<9 ; k++)
				{
					cnt+=exist(y+dy[k],x+dx[k]);
				}
				ans[cnt]++;
				visited[make_pair(y,x)]=1;
			}
		}
	}
	for(i=1 ; i<=9 ; i++)
	{
		tot-=ans[i];
	}
	printf("%lld\n",tot);
	for(i=1 ; i<=9 ; i++)
	{
		printf("%lld\n",ans[i]);
	}
}