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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
int main(){
  int a[10000][2];
  int f[10000];
  int list[5][5];
  int k[10];
  int h,w,n,i,j,x,y,z,x2,y2;
  for(i=0;i<10;i++)k[i]=0;
  scanf("%d%d%d",&h,&w,&n);
  for(i=0;i<n;i++){
    scanf("%d%d",&a[i][0],&a[i][1]);
    f[i]=0;
    if(a[i][0]-2<1||a[i][0]+2>w)f[i]|=(1<<1)+(1<<4)+(1<<7);
    if(a[i][0]-1<1)f[i]|=(1<<0)+(1<<3)+(1<<6);
    if(a[i][0]+1>w)f[i]|=(1<<2)+(1<<5)+(1<<8);
    if(a[i][1]-2<1||a[i][1]+2>h)f[i]|=(1<<3)+(1<<4)+(1<<5);
    if(a[i][1]-1<1)f[i]|=(1<<0)+(1<<1)+(1<<2);
    if(a[i][1]+1>h)f[i]|=(1<<6)+(1<<7)+(1<<8);
  }
  for(i=0;i<n;i++){
    for(j=0;j<5;j++)for(z=0;z<5;z++)list[j][z]=0;
    list[2][2]=i;
    for(j=i+1;j<n;j++){
      x=a[j][0]-a[i][0];y=a[j][1]-a[i][1];
      if(abs(x)<3&&abs(y)<3){list[2+x][2+y]=j;}
    }
    for(y=0;y<3;y++){
      for(x=0;x<3;x++){
        z=0;
        if(!(f[i]&(1<<(y*3+x)))){
          for(y2=0;y2<3;y2++){
            for(x2=0;x2<3;x2++){
              if(list[x+x2][y+y2]){
                z++;
                f[list[x+x2][y+y2]]|=1<<((2-x2)+(2-y2)*3);
              }
            }
          }
        }
        k[z]++;
      }
    }
  }
  k[0]=(h-2)*(w-2);
  for(i=1;i<10;i++)k[0]-=k[i];
  for(i=0;i<10;i++)printf("%d\n",k[i]);
}