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>
 
using namespace std;
inline void prep ()
{
  cin.tie (0);
  cin.sync_with_stdio (0);
};
const double pi = acos(-1.);
//printf(%.12lf\n,mid);
// __builtin_popcount(int) count 1's in binary
// get decimal part of number
/*
double intpart;
double thiss = modf((double)(n*m)/(double)k, &intpart);
*/
// round = (int)(num+.5)
long long mod = 1000000007;

// 32 mil = ~1 sec 
long long mymod(long long num){
  if (num< 0 && !(num%mod == 0)){
    if (mod == 1){
      return 0;
    }
    return mod-abs(num)%mod;
  }else{
    return num%mod;
  }
}
 
/*string sentence = And I feel fine...;
    istringstream iss(sentence);
*/  
 
double eps = .000000001;

int main ()
{
  prep();
  
  long long h, w, n;
  cin >> h >> w >> n;
  vector<pair<int, int> > v;
  set<pair<int, int> > have;
  for (int i=0; i<n; i++){
      int x, y;
      cin >> x >> y;
      have.emplace(x, y);
      v.emplace_back(x, y);
  }
  long long ones[10] = {0}; // don't count 0's
  for (pair<int, int> p : v){
      int x = p.first;
      int y = p.second;
      for (int i=x-2; i<x+1; i++){
          for (int j=y-2; j<y+1; j++){
              int cnt = 0;
              for (int k=0; k<9; k++){
                  int newx = i+k/3;
                  int newy = j+k%3;
                  if (newx <= 0 || newy <= 0 || newx > h || newy > w){
                      cnt = 0;
                      break;
                  }
                  if (have.count({newx, newy})){
                      cnt++;
                  }
              }
              ones[cnt]++;
          }
      }
  }
  long long full = 0;
  for (int i=1; i<=9; i++){
      ones[i] /= i;
      full += ones[i];
  }
  cout << (h-2)*(w-2)-full << '\n';
  for (int i=1; i<=9; i++){
      cout << ones[i] << '\n';
  }
  return 0; 
}