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<bits/stdc++.h>
#define LL long long
#define pii pair<LL,LL>
#define fir first
#define sec second
#define mp make_pair
using namespace std;
map<pii,LL> a;
LL X[100010],Y[100010],ans[50];
int main()
{
	LL H,W,N,tot,x,y,i;
	scanf("%lld %lld %lld",&H,&W,&N);
	for(i=1;i<=N;i++)
	{
		scanf("%lld %lld",&X[i],&Y[i]);
		a[mp(X[i],Y[i])]=1LL;
	}
	
	memset(ans,0,sizeof(ans));
	for(i=1;i<=N;i++)
	{
		x=X[i];y=Y[i];
		if(x+2<=H&&y+2<=W)
		{
			tot=a[mp(x,y)]+a[mp(x,y+1)]+a[mp(x,y+2)]+a[mp(x+1,y)]+a[mp(x+1,y+1)]+a[mp(x+1,y+2)]+a[mp(x+2,y)]+a[mp(x+2,y+1)]+a[mp(x+2,y+2)];
			ans[tot]++;
		}
		if(x+2<=H&&1<=y-1&&y+1<=W)
		{
			tot=a[mp(x,y-1)]+a[mp(x,y+1)]+a[mp(x+1,y-1)]+a[mp(x+1,y)]+a[mp(x+1,y+1)]+a[mp(x+2,y-1)]+a[mp(x+2,y)]+a[mp(x+2,y+1)]+a[mp(x,y)];
			ans[tot]++;
		}
		if(x+2<=H&&1<=y-2&&y<=W)
		{
			tot=a[mp(x,y-2)]+a[mp(x,y-1)]+a[mp(x,y)]+a[mp(x+1,y-2)]+a[mp(x+1,y-1)]+a[mp(x+1,y)]+a[mp(x+2,y-2)]+a[mp(x+2,y-1)]+a[mp(x+2,y)];
			ans[tot]++;
		}
		if(1<=x-1&&x+1<=H&&y+2<=W)
		{
			tot=a[mp(x-1,y)]+a[mp(x-1,y+1)]+a[mp(x-1,y+2)]+a[mp(x,y)]+a[mp(x,y+1)]+a[mp(x,y+2)]+a[mp(x+1,y)]+a[mp(x+1,y+1)]+a[mp(x+1,y+2)];
			ans[tot]++;
		}
		if(1<=x-1&&x+1<=H&&1<=y-1&&y+1<=W)
		{
			tot=a[mp(x-1,y-1)]+a[mp(x-1,y)]+a[mp(x-1,y+1)]+a[mp(x,y-1)]+a[mp(x,y)]+a[mp(x,y+1)]+a[mp(x+1,y-1)]+a[mp(x+1,y)]+a[mp(x+1,y+1)];
			ans[tot]++;
		}
		if(1<=x-1&&x+1<=H&&1<=y-2)
		{
			tot=a[mp(x-1,y-2)]+a[mp(x-1,y-1)]+a[mp(x-1,y)]+a[mp(x,y-2)]+a[mp(x,y-1)]+a[mp(x,y)]+a[mp(x+1,y-2)]+a[mp(x+1,y-1)]+a[mp(x+1,y)];
			ans[tot]++;
		}
		if(1<=x-2&&y+2<=W)
		{
			tot=a[mp(x-2,y)]+a[mp(x-2,y+1)]+a[mp(x-2,y+2)]+a[mp(x-1,y)]+a[mp(x-1,y+1)]+a[mp(x-1,y+2)]+a[mp(x,y)]+a[mp(x,y+1)]+a[mp(x,y+2)];
			ans[tot]++;
		}
		if(1<=x-2&&1<=y-1&&y+1<=W)
		{
			tot=a[mp(x-2,y-1)]+a[mp(x-2,y)]+a[mp(x-2,y+1)]+a[mp(x-1,y-1)]+a[mp(x-1,y)]+a[mp(x-1,y+1)]+a[mp(x,y-1)]+a[mp(x,y)]+a[mp(x,y+1)];
			ans[tot]++;
		}
		if(1<=x-2&&1<=y-2)
		{
			tot=a[mp(x-2,y-2)]+a[mp(x-2,y-1)]+a[mp(x-2,y)]+a[mp(x-1,y-2)]+a[mp(x-1,y-1)]+a[mp(x-1,y)]+a[mp(x,y-2)]+a[mp(x,y-1)]+a[mp(x,y)];
			ans[tot]++;
		}
	}
	tot=0ll;for(i=1;i<=9;i++){ans[i]/=i;tot+=ans[i];}
	ans[0]=(H-2LL)*(W-2LL)-tot;
	for(i=0;i<=9;i++)printf("%lld\n",ans[i]);
	return 0;
}