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 <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const LL base=(int)(1e9)+7;
const int maxn=1e5+10;
int h,w,n,tot;
LL a[maxn<<4],ans[20];
int main() {
    while (~scanf("%d%d%d",&h,&w,&n)) {
        memset(ans,0,sizeof(ans));
        tot=0;
        LL zero=(1LL*h-2)*(w-2);
        int x,y;
        for (int i=1;i<=n;i++) {
            scanf("%d%d",&x,&y);
            for (int j=0;j<3;j++) for (int k=0;k<3;k++)
                if (x-j>=1 && x-j<=h-2 && y-k>=1 && y-k<=w-2)
                    a[++tot]=base*(x-j)+(y-k);
        }
        sort(a+1,a+tot+1);
        int cnt=1;
        for (int i=1;i<=tot;i++) {
            if (a[i]==a[i+1]) cnt++;
            else ans[cnt]++, cnt=1, zero--;
        }
        printf("%lld\n", zero);
        for (int i=1;i<=9;i++) printf("%lld\n", ans[i]);
    }
    return 0;
}