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: Wrong Answer)

#include <cstdio>
#include <cstring>
#include <set>
#define f first
#define s second
#define par pair<int,int>
using namespace std;

typedef long long llint;
const int MAXN = 100005;

int h, w, n, x,y;
llint ans[10];
bool around[5][5];

set < pair<int,int> > s;

bool check(int x, int y) {
  return x < 0 || x >= h || y < 0 || y >= w;
}

void count(int x, int y) {
  for (int i = 0; i < 3; ++i) {
    for (int j = 0; j < 3; ++j) {
      int cnt = 0;
      bool flag = 1;
      for (int ii = 0; ii < 3; ++ii)
        for (int jj = 0; jj < 3; ++jj) {
          if (check(x+i-2+ii, y+j-2+jj)) {
            flag = 0;
          }
          cnt += around[i+ii][j+jj];
        }
      if(flag) {
        ans[cnt]++;
      }
    }
  }
}

void print() {
  for (int i = 0; i < 5; ++i) {
    for (int j = 0; j < 5; ++j)
      printf("%d", around[i][j]);
    printf("\n");
  }
}

int main() {
  scanf("%d%d%d", &h, &w, &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &x, &y); x--; y--;
    s.emplace(x, y);
  }
  for (auto it = s.begin(); it != s.end(); it++) {
    memset(around, 0, sizeof around);
    for (int ii = -2; ii < 3; ++ii) {
      for (int jj = -2; jj < 3; ++jj) {
        if (s.find(par(it->f+ii, it->s+jj)) != s.end()) {
          around[ii+2][jj+2] = 1;
        }
      }
    }
    printf("%d %d\n", it->f, it->s);
    print();
    count(it->f, it->s);
  }
  ans[0] += (llint)(h-2) * (w-2);
  for (int i = 1; i < 10; ++i) {
    ans[i] /= i;
    ans[0] -= ans[i];
  }
  for (int i = 0; i < 10; ++i)
    printf("%lld\n", ans[i]);
  return 0;
}