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: Accepted)

#include<stdio.h>
long long R=1,C=1,H[2000010],N[2000010];
//評価関数（いまはMAX）
int hyouka(int a,int b){
  if(C<b)return 1;
  if(C<a||b==0)return 0;
  return N[H[a]]<N[H[b]]?1:0;
}
//挿入関数
void hin(long long a){
  int i=C++;
  for(N[H[0]=R]=a;hyouka(0,i/2);i/=2)H[i]=H[i/2];
  H[i]=R++;
}
//取り出す関数
int hout(){
  int rt=H[1],i,j=2,k=H[--C];
  for(i=1;hyouka(i,C);i=j)H[i]=H[j=i*2+1-hyouka(i*2,i*2+1)];
  H[j/2]=k;
  return rt;
}
int Y[]={1,1, 1,0,0, 0,-1,-1,-1};
int X[]={1,0,-1,1,0,-1, 1, 0,-1};
int main(){
  int h,w,n,i,j,ny,nx;
  long long a,b,c[10]={0};
  scanf("%d %d %d",&h,&w,&n);//printf("%d %d %d\n",h,w,n);
  for(i=0;i<n;i++){//printf("%d\n",i);
    scanf("%lld %lld",&a,&b);
    for(j=0;j<9;j++){
      ny=a+Y[j];
      nx=b+X[j];
      if(ny>=h||ny<2||nx>=w||nx<2)continue;//printf("%d %d\n",ny-1,nx-1);
      hin((ny-1LL)*w+nx-1);
    }//printf("\n");
  }//printf("a\n");
  if(C-1){
    b=N[hout()];
    for(j=1;C-1;){//printf("%lld %lld\n",b/w,b%w);
      a=N[hout()];
      if(b==a)j++;
      else{
	c[j]++;
	j=1;
      }
      b=a;
    }
    c[j]++;
  }
  c[0]=(h-2LL)*(w-2);
  for(i=1;i<10;i++)c[0]-=c[i];
  for(i=0;i<10;i++)printf("%lld\n",c[i]);
  return 0;
}