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;

typedef long long ll;

const int INF = 1e9;
const ll LLINF = 1e18;
const ll MOD = (ll)1e9 + 7;

int main()
{
  int H, W, N; scanf("%d %d %d", &H, &W, &N);
  int a, b;
  vector<pair<int, int>> V;
  for (int i = 0; i < N; ++i) {
    scanf("%d %d", &a, &b);
    for (int dr = -2; dr <= 0; ++dr) {
      for (int dc = -2; dc <= 0; ++dc) {
        int nr = a + dr, nc = b + dc;
        if (1 <= nr and nr + 2 <= H and 1 <= nc and nc + 2 <= W) {
          V.push_back(make_pair(nr, nc));
        }
      }
    }
  }
  vector<ll> ans(10, 0);
  sort(V.begin(), V.end());
  int n = V.size();
  for (int i = 0, j; i < n; i = j) {
    j = i;
    while (j < n and V[j] == V[i]) { ++j; }
    int cnt = j - i;
    ++ans[cnt];
  }
  ans[0] = 1LL* (H - 2) * (W - 2) - accumulate(ans.begin(), ans.end(), 0);
  for (int i = 0; i < 10; ++i) {
    printf("%lld\n", ans[i]);
  }
  return 0;
}