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 <iostream>
#include <vector>
#include <utility>
#include <set>
#include <array>
#include <numeric>

using namespace std;

using Pos = pair<uint64_t, uint64_t>;

uint64_t H, W, N;
set<Pos> Blacks;
array<uint64_t, 10> Spots;


inline uint64_t count(uint64_t a, uint64_t b)
{
  uint64_t ret = 0;
  for (uint64_t i=a; i<a+3; ++i)
    for (uint64_t j=b; j<b+3; ++j)
      ret += Blacks.count(make_pair(i, j));

  return ret;
}

int main()
{
  cin >> H >> W >> N;
  for (uint64_t i=0; i<N; ++i) {
    uint64_t a, b;
    cin >> a >> b;
    Blacks.insert(make_pair(a, b));
  }

  for (const auto& p : Blacks) {

    uint64_t min_a = p.first - 2 < 1 ? 1 : p.first - 2;
    uint64_t max_a = H < p.first + 2 ? H - 2 : p.first;
    uint64_t min_b = p.second - 2 < 1 ? 1 : p.second - 2;
    uint64_t max_b = W < p.second + 2 ? W - 2 : p.second;

    for (uint64_t a=min_a; a<=max_a; ++a)
      for (uint64_t b=min_b; b<=max_b; ++b)
        ++Spots[count(a, b)];
  }

  for (uint64_t i=1; i<10; ++i)
    Spots[i] /= i;

  Spots[0] = (H-2) * (W-2) - accumulate(Spots.cbegin()+1, Spots.cend(), static_cast<uint64_t>(0));

  for (uint64_t i=0; i<10; ++i)
    cout << Spots[i] << endl;

  return 0;
}